Sum of a Triple

Since $a,b,c \geq 1$ , it follows that $\lg a, \lg b, \lg c \geq 0$ . Take log at both sides of the equality given: $abc = 100$ , $\lg abc = \lg 100$ , $\lg a + \lg b + \lg c = 2 \quad (I)$

Take log at both sides of the inequality given (we can do that because the base of the log (10) is greater than 1 hence is an increasing function, and the log function is injective) : $a^{\lg a} b^{\lg b} c^{\lg c} \geq 10000$ $\lg (a^{\lg a} b^{\lg b} c^{\lg c}) \geq \lg 10000$ , $\lg a^{\lg a} + \lg b^{\lg b} + \lg c^{\lg c} \geq 4$ $(\lg a)^2 + (\lg b)^2 + (\lg c)^2 \geq 4 \quad (II)$

Squaring both sides of (I) leads to: $(\lg a)^2 + (\lg b)^2 + (\lg c)^2 + 2(\lg a \lg b + \lg a \lg c + \lg b \lg c) = 4$ , $(\lg a)^2 + (\lg b)^2 + (\lg c)^2 = 4 - 2(\lg a \lg b + \lg a \lg c + \lg b \lg c)$ . Compare that to (II) and it follows that: $4 - 2(\lg a \lg b + \lg a \lg c + \lg b \lg c) \geq 4$ $- 2(\lg a \lg b + \lg a \lg c + \lg b \lg c) \geq 0$ , $\lg a \lg b + \lg a \lg c + \lg b \lg c \leq 0$ But $\lg a, \lg b, \lg c \geq 0$ , therefore $\lg a \lg b + \lg a \lg c + \lg b \lg c \geq 0$ , which implies: $\lg a \lg b + \lg a \lg c + \lg b \lg c = 0$ $\lg a, \lg b$ and $\lg c$ can't be all equal to zero, since $\lg a + \lg b + \lg c = 2$ . Thus, two numbers among $\lg a, \lg b, \lg c$ are equal to zero and the other is equal to 2. Without loss of generality, let $\lg a = \lg b = 0$ and $\lg c = 2$ . It follows that $a = b = 1$ and $c = 100$ , therefore $a + b + c = 102$ .

[Punctuation edits - Calvin]

We take the $\lg$ in the first equation and we get $\lg a+\lg b+\lg c=2$ In the second equation we write $a,b,c$ as $10^{\lg a}$ etc, so we have $10^{\lg^2a+\lg^2b+\lg^2c}\ge10^4$ , which means $\lg^2a+\lg^2b+\lg^2c\ge4$ But $\lg^2a+\lg^2b+\lg^2c=\\=(\lg a+\lg b+\lg c)^2-2(\lg a\lg b+\lg b\lg c+\lg c\lg a)$ . Hence we have, substituting the sum and dividing by $-2$ , $\lg a\lg b+\lg b\lg c+\lg c\lg a\le0$ But since $\lg x\ge0 \ \forall x\ge1$ , we must have WLOG $\lg a=\lg b=0$ , i.e. $(a,b,c)=(1,1,100)$

Let $a=10^{x}, b=10^{y}, c=10^{z},$ so that $x, y, z\ge 0$ . We know that $abc=10^{x+y+z}=100,$ so $x+y+z=2.$ Also that $a^{\lg a}=10^{x\lg 10^x}=10^{x^2},$ so $b^{\lg b}=10^{y^{2}}$ , $c^{\lg c}=10^{z^{2}}$ , Therefore, $a^{\lg a} b^{\lg b} c^{\lg c}=10^{x^{2}+y^{2}+z^{2}}\ge 10000.$ So $x^{2}+y^{2}+z^{2}\ge 4$ .

Now $4=2^{2}=(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2yz.$ Since $x^{2}+y^{2}+z^{2}\ge 4, 2xy+2xz+2yz\le 0.$ But since $x,y,z\ge 0,$ the equality must hold, i.e. $2xy+2xz+2yz=0,$ so two of $x,y,z$ must be $0$ , and the other one must be $2$ .

So $\{x,y,z\}=\{2,0,0\},$ $\{a,b,c\}=\{100,1,1\}.$ Hence in any case $a+b+c=100+1+1=102.$ Q.E.D.

Let $x = \log {10} a$, $y = \log {10} b$, and $z = \log {10} c$. Then $a = 10^x$, $b = 10^y$, and $c = 10^z$. The inequalities $a$, $b$, $c \ge 1$ tell us that $x$, $y$, $z \ge 0$, and the given equations $abc = 100$ and $a^{\log {10} a} b^{\log {10} b} c^{\log {10} c} \ge 10000$ tell us that $10^{x + y + z} = 100$ and $10^{x^2 + y^2 + z^2} \ge 10000$, so $x + y + z = 2$ and $x^2 + y^2 + z^2 \ge 4$.

Since $x$, $y$, $z \ge 0$ and $x + y + z = 2$, we have that $0 \le x \le 2$. Similarly, $0 \le y \le 2$ and $0 \le z \le 2$. Hence, $x(2 - x) \ge 0, \quad y(2 - y) \ge 0, \quad z(2 - z) \ge 0. \quad (*)$ Adding all these inequalities, we get $2x + 2y + 2z \ge x^2 + y^2 + z^2.$ Then $x^2 + y^2 + z^2 \le 2(x + y + z) = 4$. But $x^2 + y^2 + z^2 \ge 4$, so we must have $x^2 + y^2 + z^2 = 4$, and we must have equality in each inequality in $(*)$. This means each of $x$, $y$, and $z$ is equal to 0 or 2. Since $x + y + z = 2$, two of $x$, $y$, and $z$ must be equal to 0, and the third must be equal to 2. Then two of $a$, $b$, and $c$ must be equal to $10^0 = 1$, and the third must be equal to $10^2 = 100$, so $a + b + c = 1 + 1 + 100 = 102$.

Taking lg of the first equation gives us $lg\ a + lg\ b + lg\ c = lg\ 100 = 2$ and squaring gives us $(lg\ a)^2+(lg\ b)^2+(lg\ c)^2 + 2(lg\ a)(lg\ b) + 2(lg\ b)(lg\ c) + 2(lg\ a)(lg\ c) = 4$ . Taking lg of the second equation gives us $(lg\ a)^2+(lg\ b)^2+(lg\ c)^2 = 4$ and subtracting the two equations gives $2(lg\ a)(lg\ b) + 2(lg\ b)(lg\ c) + 2(lg\ a)(lg\ c) = 0$ .

Note that $a,b,c \geq 1 \Rightarrow lg\ a,lg\ b,lg\ c \geq 0$ . If at least 2 of $lg\ a, lg\ b, lg\ c$ are positive then $2(lg\ a)(lg\ b) + 2(lg\ b)(lg\ c) + 2(lg\ a)(lg\ c) > 0$ since at least 1 term is positive and the others are non-negative. They also can't all be 0 so the only possibility is that exactly 2 are 0 and since their sum is 2 the third must be 2. Therefore, $a+b+c = 10^{lg\ a}+10^{lg\ b}+10^{lg\ c} = 10^2+10^0+10^0 = 102$

Let x = lg(a), y = lg(b), and z = lg(c). Note that:

$a = 10^{lg(a)} = 10^x$

$b = 10^y$

$c = 10^z$

Thus, the first equation gives:

$100 = abc = 10^x 10^y 10^z = 10^{x+y+z}$

Taking both sides log base 10:

$2 = x + y + z$

$4 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$

From the second equation:

$10000 \le a^{lg(a)} b ^{lg(b)} c^{lg(c)} = (10^{lg(a)})^{lg(a)} (10^{lg(b)})^{lg(b)} (10^{lg(c)})^{lg(c)}$

$10000 \le (10^x)^x (10^y)^y (10^z)^z = 10^{x^2 + y^2 + z^2}$

$4 \le x^2 + y^2 + z^2$

Subtracting these two equations yields:

$0 \ge 2(xy+ yz+zx)$

$0 \ge xy + yz + zx$ (*)

Note that since $a,b,c \ge 1$ , $x,y,z \ge 0$ . Thus, for (*) to be true, every term must be 0. This means that at least two of x, y, z must be 0. WLOG let x = 0 and y = 0. Then, z = 2 - x - y = 2. Thus, a = 1, b = 1, c = 100, so:

$a+b+c = 1 + 1 +100 = \boxed{102}$

(1) The given inequality, with a bit of algebra (change $x^{\lg x}$ to $10^{\lg^2 x}$ and collect the powers), becomes $\lg^2 a+\lg^2 b+\lg^2 c≥4$ .

(2) The given equality is $\lg a + \lg b + \lg c = 2$ .

Substitute $\lg a, \lg b, \lg c$ by $x, y, z$ respectively. Notice that now (1) describes a sphere and (2) is a plane. Notice also that we must have $x, y, z \geq 0$ to satisfy $a, b, c \geq 1$ . This is a trivial problem, though a drawing might be helpful.

The only points $(x, y, z)$ that satisfy the constraints are permutations of $(2, 0, 0)$ . Thus $a=100, b=1, c=1$ and $a+b+c=102$ .

We rename $a = 10^x$ , $b = 10^y$ and $c = 10^z$ . Then $x = \lg a$ , $y = \lg b$ and $z = \lg c$ . Also, $x, y, z \geq 0$

The conditions can be written: $abc = (10^x)(10^y)(10^z) = 100$ , $a^{\lg a} b^{\lg b} c^{\lg c} = (10^x)^x (10^y) ^y (10^z)^z \geq 10000$

Simplifying: $10^{x + y + z} = 10^2$ , $10^{x^2 + y^2 + z^2} \geq 10^4$

Then, we have: $x + y + z = 2$ , $x^2 + y^2 + z^2 \geq 4$

If we square the first expression, then we have: $x + y + z + 2(xy + yz + zx) = 4$ , $x + y + z = 4 - 2(xy + yz + zx)$

But $x^2 + y^2 + z^2 \geq 4$ and $x + y + z = 4 - 2(xy + yz + zx)$ so: $4 - 2(xy + yz + zx) \geq 4 \Rightarrow 4 - 4 \geq 2(xy + yz + zx) \Rightarrow$ $0 \geq 2(xy + yz + zx) \Rightarrow 0 \geq xy + yz + zx$

Also, $x, y, z \geq 0$ , then $xy = 0, yz = 0, zx = 0$ .

Wlog, suppose $x \leq y \leq z$ . If $x = 0$ but $y, z \neq 0$ then $yz \neq 0$ , $x = y = 0$ , and $z = 2$ .

But $a = 10^x = 10^0 = 1, b = 10^y = 10^0 = 1$ and $c = 10^z = 10^2 = 100$ , then $a + b + c = 1 + 1 + 100 = 102$ .

Knowing that if $x\geq y\geq 1$ than $\log x \geq \log y$ Taking log from both sides of both the equations we get: $\begin{cases} \log a + \log b + \log c=2 \\ (\log a)^2+(\log b)^2+(\log c)^2 \geq 4 \\ \end{cases}$ the second equation becomes: $(\log a + \log b + \log c)^2-2(\log a \log b+\log b \log c+\log a \log c) \geq 4$ using the first equation we get: $-2(\log a \log b+\log b \log c+\log a \log c) \geq 0$ $(\log a \log b+\log b \log c+\log a \log c) \leq 0$ Knowing that any log has to be positive we get: $(\log a \log b+\log b \log c+\log a \log c) = 0$ This means that at least two of the logs are $0$ . Knowing that the sum of the logs has to be $2$ we have that the third log equals $2$ .

Therefore the three numbers are: $1$ , $1$ and $100$ . And their sum is: $\fbox{102}$

$\begin{array}{l} a^{\lg a} b^{\lg b} c^{\lg c} \ge 10000 \\ 10^{(\lg a)^2 + (\lg b)^2 + (\lg c)^2 } \ge 10000 \\ \left( {\lg a} \right)^2 + \left( {\lg b} \right)^2 + \left( {\lg c} \right)^2 \ge 4 \\ \end{array}$ again $abc = 100$ $\lg a + \lg b + \lg c = 2$ Now $\left( {\lg a + \lg b + \lg c} \right)^2 = \left( {\lg a} \right)^2 + \left( {\lg b} \right)^2 + \left( {\lg c} \right)^2 + 2\left( {\lg a.\lg b + \lg b.\lg c + \lg c.\lg a} \right)$ $\left( {\lg a} \right)^2 + \left( {\lg b} \right)^2 + \left( {\lg c} \right)^2 = 4 - 2\left( {\lg a.\lg b + \lg b.\lg c + \lg c.\lg a} \right)$ $\left( {\lg a} \right)^2 + \left( {\lg b} \right)^2 + \left( {\lg c} \right)^2 \le 4$ hence among $\lg a,\lg b{\rm{ }}and\lg c$ two of them are zero hence $a + b + c = 102$

From $abc=100 \Rightarrow \lg{a}+\lg{b}+\lg{c}=2 \Rightarrow$ $\lg{a}^2+\lg{b}^2+\lg{c}^2+2\lg{a} \lg{b}+2\lg{a} \lg{c}+2\lg{b} \lg{c}=4$

From $a^{\lg{a}} b^{\lg{b}} c^{\lg{c}} \geq 10000 \Rightarrow$ $\lg{a}^2+\lg{b}^2+\lg{c}^2 \geq 4$ that is $4+2\lg{a} \lg{b}+2\lg{a} \lg{c}+2\lg{b} \lg{c} \leq 4 \Rightarrow$ $2\lg{a} \lg{b}+2\lg{a} \lg{c}+2\lg{b} \lg{c} \leq 0$

But $a, b, c \geq 1 \Rightarrow \lg{a}, \lg{b}, \lg{c} \geq 0$

The inequality $2\lg{a} \lg{b}+2\lg{a} \lg{c}+2\lg{b} \lg{c} \leq 0$ is true when all terms $\lg{a} \lg{b}, \lg{a} \lg{c}, \lg{b} \lg{c}$ are zero i.e. at least two from $\lg{a}, \lg{b}, \lg{c}$ are equal with zero. That means that if we take $\lg{b}=\lg{c}=0$ then $b=c=1$ and because $abc=100$ , $a$ must be equal with $100$ .

Of course same result if $a,b,c$ are interchanged. So the final result is $a+b+c=102$ .

Let $x = \lg a$ , $y = \lg b$ , and $z = \lg c$ . Then

$\lg abc\ = x + y +z = 2$ , and

$\lg {a^{\lg a} b^{\lg b} c^{\lg c}} = x^2 + y^2 + z^2 \geq 4$ .

$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) = 4$ , so either

$x^2 + y^2 + z^2 = 4$ and $xy + xz + zy = 0$ or

$x^2 + y^2 + z^2 > 4$ and $xy + xz + zy < 0$ .

But $a, b, c \geq 1$ , so $x, y, z \geq 0$ and $xy + xz + yz \geq 0$ . For $xy + xz + yz$ to equal $0$ , all three terms must be $0$ which is only possible if at least two of $x, y, z$ are $0$ . So let $x, y = 0$ in which case $a, b = 1$ and $c = 100$ . $1 + 1 + 100 = 102$ .

Assume WLOG that a>=b>=c.

The first equation states, abc = 100. Solving for a, we have a=100/(bc). Since a,b,c >= 1, we have that a <= 100. Similarly, if we solved for b and c, we'd also have b <= 100 and c <= 100.

Now divide the second equation by the first equation twice. We have a^(lg a - 2) b^(lg b - 2) c^ (lg c - 2) >= 1. Note that in order for the product of 3 factors to be at least 1, at least 1 of the factors must be at least 1. (in other words, if xyz >= 1, at least one of x>= 1, y>=1, or z>=1 must be true)

Since we assumed at the beginning that a is biggest (a >=b>=c ), we must have that a^(lg a - 2) >=1. Since a >= 1, either lg a - 2 >= 0 or a = 1. In the first case, a >= 100, which means a=100, and b=c=1. In the second case, since a>=b>=c , a=b=c=1, which is not possible because 1 1 1 is not 100. Therefore the only possible triple (ordered by size) is (100,1,1), and the sum is 100+1+1=102.

from abc=100: abc=100 => lg(a)+lg(b)+lg(c)=2

from a^lg(a)b^lg(b)c^lg(c)>=10000: lg(a^lg(a)b^lg(b)c^lg(c))>=4 lg(a^lg(a)) + lg(b^lg(b) + lg(c^lg(c)) >= 4 lg(a)^2 + lg(b)^2 + lg(c^2) >= 4 (lg(a) + lg(b) + lg(c))^2 - 2(lg(a)lg(b)+lg(a)lg(b)+lg(b)lg(c)) >=4 4 - 2(lg(a)lg(b)+lg(alg(b)+lg(b)lg(c)) >= 4 (lg(a)lg(b)+lg(a)lg(c)+lg(b)lg(c)) <= 0 because lg(a),lg(b),lg(c) are non negative, (lg(a)lg(b)+lg(a)lg(b)+lg(b)lg(c)) = 0 no more then one of a,b,c can be >1, otherwise one of the addends is greater than 0. So a=b=1 and c=100

Let a=10^x, b=10^y, c=10^z NOW USING ABC=100! WE GET X+Y+Z=2 ALSO A,B,C>=1 so x,y,z>=0 Also equation 2 given in question lead to x^2+y^2+z^2=4 So now (x+y+z)^2is also 4 so xy+yz+zx=0 With all non negative so two among three are 0 So letx=0,y=0,z=2 Then A=1 B=1 C=100 so A+B+C= 102

Let $A= \lg a, B = \lg b, C = \lg c$ . Since $a, b, c \geq 1$ , we have $A, B, C \geq 0$ . Since $abc = 100$ , taking log base 10 on both sides gives $A+B+C = 2$ . Since $a^{\lg a} b^{\lg b} c^{\lg c} \geq 10000$ , taking log base 10 on both sides gives $\lg (a^{\lg a}) + \lg( b^{\lg b} ) + \lg (c^{\lg c}) \geq 4 \Rightarrow A^2 + B^2 + C^2 \geq 4$ .

Then $A^2 + B^2 + C^2 \geq 4 = (A + B + C)^2$ $\Rightarrow A^2 + B^2 + C^2 \geq A^2 + B^2 + C^2+ 2AB + 2AC + 2BC$ $\Rightarrow AB + AC + BC \leq 0$ . But since $AB, BC$ and $CA$ are all non-negative, $AB+BC+CA \geq 0$ . Hence, $AB+BC+CA = 0$ , which implies that $AB=0, BC=0, CA=0$ . Thus at least 2 of $A, B$ and $C$ are 0. Without loss of generality, we set $B=C=0$ , which gives $A = 2 - 0 - 0 = 2$ . Thus $a = 10^2 = 100, b = 10^0 = 1, c = 10^0 = 1$ , and $a+b+c = 102$ .

Since it says is greater than or equal to 1 (this is very uncommon - usually its strictly greater than) immediately I think of 'lets see what happens when a and b are equal to 1'. -> thus c = 100

I worked it out, and it worked :D

Get the logarithmic expression equivalent to the first equation as well as the logarithmic expression equivalent to the second (this is valid because the left and the right side of the inequality are both equal or greater than one.) Then, $\begin {cases} \lg a +\lg b = 2- \lg c, \\ (\lg a)^2 + (\lg b)^2 + (\lg c)^2 \geq 4 \\ \end {cases}$ .

Squaring the first equation yields $(\lg a)^2 + (\lg b)^2-2 \cdot (\lg a) \cdot (\lg b)= 4$ . Applying this to the second equation yields $2 \cdot (\lg a) \cdot (\lg b) \geq 0$ . Then, $10^ {{\lg a}{\lg b}} \geq 1$ , or equivalently $ab \geq 1$ . But under the assumption that a,b, and c are all greater than or equal to one, then it follows that both a and b are 1, and so c=100. Therefore, * a + b+ c = 102 *.

An important observation is that $10000=100^2$ . This gives us the idea to set $a=100$ so that $b,c=1$ , and $a^{\lg a}b^{\lg b}c^{\lg c}=100^2 \times 1^0 \times 1^0=10000.$

To show that there are no other solutions (except for by symmetry), we consider a value for a less than 100 which we shall call $100-m$ . We must then change the value(s) of b, c, or both. We begin with changing only the value of b: To satisfy $abc=100$ , we have $a=100-m$ and $b=\frac{100}{100-m}$ . (c is still equal to 1.) Thus we have $a^{\lg a}b^{\lg b}c^{\lg c}=(100-m)^{\lg (100-m)}(\frac{100}{100-m})^{\lg \frac{100}{100-m}}(1)=$ $(100-m)^{\lg (100-m)}(\frac{(\frac{100}{100-m})^{\lg 100}}{(\frac{100}{100-m})^{\lg (100-m)}})=(\frac{100}{100-m})^{2}<100^2$ Therefore, if we "transfer" some amount m away from a and increase b (or c by symmetry), our conditions are not met. This can be shown similarly when a is decreased and b and c are both increased. Therefore, our only solution is $(a,b,c)=(100,1,1)$ , or permutations. This gives us our answer of $100+1+1=\boxed{102}$ .

The key is that a b c=100

So a,b,c can be

1. 100 1 1

2. 50 2 1

3. 25 4 1 or 25 2 2

4. 20 5 1

5. 10 10 1 or 10 5 2

However,

• 50^{log(50)} is about 770,

• 25^log(25) is about 90,

• 2^log(2) is about 1.2,

• 5^log(5) is about 3

• 10^log(10) is 10

• 4^log(4) is about 2

• 1^log(1) is 1

• 100^log(100) is 10000 !

So the only combination of a, b and c that satisfies the second restriction is

• 100 1 1 !!!

Rewrite a^lg(a)b^lg(b)c^lg(c)>=10000 as a^lg(a)b^lg(b)c^lg(c)>=100^lg(100), because 10000=100^2=100^lg(100). Since a^lg(a) and b^lg(b) are 1 when a=1 and b=1, the solution is completed when c=100, yielding a+b+c = 102.

assume WLOG that $a\geq b\geq c\geq 1$ , then $\lg a\ge\lg b\ge lg c\ge 0$ .

Then $(abc)^{\lg a}\geq a^{\lg a} b^{\lg b} c^{\lg c}\geq 10,000$ , which implies that $100^{\lg a}\geq 10,000$ , and so $\lg a\geq 2$ . Thus $a\geq 100$ , and a=100, b=1, c=1, so the sum is $\boxed{102}$

Think about this geometrically.

We agree on two facts:

(log a)^2 + (log b)^2 + (log c)^2 >= 4 and log a + log b + log c = 2

let log a = x, log b = y, and log c = z. note: x, y, z >= 0.

The first inequality describes all points outside a sphere of radius 2 (=sqrt(4)). The second equation defines a plane passing through the points (0, 0, 2), (0, 2, 0) and (2, 0, 0). These three points are the only points on the plane which are 2 units away from the origin. All other points on the plane (assuming x, y, and z >= 0) are closer.

The corresponding values of a, b, and c at these three points are (1, 1, 100), (1, 100, 1) and (100, 1, 1).

a + b + c =102.

Observe that $x^{log x}$ is smaller than $x$ if $x$ is a positive divisor of 100 which is less than 10.

Without loss of generality let $a \geq b \geq c$ . Then observe that because $abc=100$ , at most 2 of $a,b,c$ are larger than 10.

This only occurs when $a=b=10$ . It is easy to check that this does not satisfy the problem requirements.

Now we know that only $a \geq 10$ , it suffices to check $a=20,25,50,100$ to conclude that $a=100$ .

But there is a quicker way, in order for the answer to be unique we know immediately that only $a=100$ can work because as $a$ increases so does the overall product. (Also it is clear that none of the sums are the same)

So wrapping up, we see that $a=100, b=1, c=1$ and $a+b+c=102$

abc = 100 * 1, consume that ab=100 and c=1

Then, log a + log b = 2, log b = 2 - log a ..................(1)

(log a)^2 + (log b)^2 + (log c)^2 = 4

Since c = 1, log c = log 1 = 0 So that,

(log a)^2 + (log b)^2 = 4 ...............(2)

Substitute equation (1) into equation (2), we get:

(log a)^2 + (2-log a)^2 = 4 (log a)^2 + 4-4 log a + (log a)^2 = 4 (log a)^2 - 2 log a = 0

Factorize it and we get a = 100 or a = 1 Take a = 1 then substitute its value into equation (1) and we get the value b = 100.

Then, a + b + c = 1 + 100 + 1 = 102