Find Ed's numbers

Since $a,b,c$ are positive reals, so are $a^4, b^4, c^4$ .

By AM-GM, we have: $\dfrac{a^4+b^4+c^4+16}{4} \ge \sqrt[4]{a^4 \cdot b^4 \cdot c^4 \cdot 16} = 2abc$ ,

or equivalently: $a^4+b^4+c^4+16 \ge 8abc$ .

In AM-GM, equality occurs iff all the numbers are equal, i.e. $a^4 = b^4 = c^4 = 16$ .

Since $a,b,c$ are positive reals, we have $a = b = c = 2$ , and thus, $abc = \boxed{8}$ .

Knowing that $AM \ge GM$ , and applying that in the LHS, we get

$\dfrac{a^4 + b^4 + c^4 + 2^4}{4} \ge \sqrt[4]{(abc2)^4}$ $\rightarrow a^4 + b^4 + c^4 + 2^4 \ge 8abc$

Since the equality holds, we can say that $a=b=c=2$ . Then $abc = 2\times 2\times 2 = 8$ .

Consider the function

$f(a,b,c) = a^4+b^4+c^4+16-8abc$ ,

so that we want to solve $f(a,b,c) = 0$ . Note that $a=b=c=2$ is a solution. I will prove that this is an absolute minimum, i.e. that $f(a,b,c)$ is positive for all $(a,b,c) \neq (2,2,2)$ , from which it follows that this solution is the only one.

First note that $f$ is differentiable on the entire domain, and that $f$ tends to positive infinity as $a$ , $b$ or $c$ goes to infinity. The absolute minimum is thus to be found either at a boundary of the domain or at a critical point.

At the boundary (technically, we have an open boundary since $a$ , $b$ and $c$ are not allowed to be zero, but "closing the boundary" will not alter the final conclusion), we have for instance $a=0$ , which gives $f(a,b,c) = b^4+c^4+16$ , which is obviously at least 16, hence not an absolute minimum (since we've already found $f(2,2,2) = 0$ ). The same of course applies to $b=0$ and $c=0$ .

We are thus left with critical points, for which $df/da = df/db = df/dc = 0$ . This gives

$0 = 4a^3 - 8bc = 4b^3 - 8ca = 4c^3 - 8ab$ ,

from which we establish $a^3 = 2bc$ , $b^3 = 2ca$ and $c^3 = 2ab$ . Multiplying the equations by $a$ , $b$ and $c$ respectively, we get

$a^4 = b^4 = c^4 = 2abc$ ,

with the only solutions $a=b=c=0$ and $a=b=c=2$ . The latter is thus the absolute minimum, which concludes the proof. The only solution to the original equation is thus $a=b=c=2$ with $abc = \boxed{8}$ .

We are given $a^4 + b^4 + c^4 + 2^4 = 4(2abc)$ . Cheesy way out: note that $a=b=c=2$ works giving the answer of 8. Another way: by AMGM, if they are all positive,

$\frac{a^4 + b^4 + c^4 + 2^4}{4} \geq \sqrt[4]{2^4a^4b^4c^4}$

with equality iff $a=b=c=2$ .

By the AM-GM inequality, $a^4 + b^4 + c^4 + 2^4 \ge 4 (2^4 a^4 b^4 c^4)^{1/4} = 8abc,$ with equality attained when $a = b = c = 2$ . Checking, we find $3(2^4) + 16 = 64 = 8(2^3),$ so $abc = 8$ .

Notice by AM-GM, $\dfrac{a^4 + b^4 + c^4 + 16}{4} \ge \sqrt[4]{16a^4b^4c^4} = 2abc$ , so $a^4 + b^4 + c^4 + 16 \ge 8abc$ . Equality occurs when $a^4 = b^4 = c^4 = 16$ , so $abc = \boxed{8}$ .

Using the AM-GM inequality : $AM(a^4, b^4, c^4, 2^4) \geqslant GM(a^4, b^4, c^4, 2^4)$ $\frac{1}{4}(a^4 + b^4 + c^4 + 2^4) \geqslant (a^4 b^4 c^4 2^4)^{\frac{1}{4}}$ $\frac{1}{4}(a^4 + b^4 + c^4 + 2^4) \geqslant 2abc$

The equality case $AM = GM$ , which is the original equation, only occurs when all terms are equal, i.e. $2 = a = b = c$ giving $abc = \boxed{8}$ .

$AM \geq GM$

$\frac {a^4+b^4+c^4+16}{4}\geq\sqrt[4]{16a^4b^4c^4}$

$a^4+b^4+c^4+16\geq8abc$

and LHS=RHS if $a^4=b^4=c^4=16$

$a=b=c=2$

so $abc$ = 8

We can rewrite the given equation as $16+\sum a^4 = 4(2abc).$ However, by the AM-GM inequality we have $$\frac{1}{4}\left(2^4+\sum a^4\right)\ge\sqrt[4]{2^4+\sum a^4},$$ and equality occurs if and only if $a=b=c=2$ , so $abc=\boxed{8}$

The solution is easily obtained using AM-GM inequality, and using the fact that equality holds when the quantities are equal to each other.

Here, taking AM GM inequality, we have

a^4 + b^4 + c^4 + 16 >= 8abc which is in fact, an equality here and therefore a^4 = b^4 = c^4 = 16. As we want positive values of a, b, c, we infer a=b=c = 2.

So abc = 8

Note that by AM-GM inequality, the expression on the LHS is always greater than or equal to the RHS Equality only occurs when a^{4}=b^{4}=c^{4}=16, that is a=b=c=2, hence abc=8

We know that for positive reals a,b,c,d we have the inequality a^{4} + b^{4} + c^{4} + d^{4} >= 4abcd putting d=2, we have a^{4} + b^{4} + c^{4} + 2^4 >=4.a.b.c.2 or, a^4 + b^4 + c^4 + 16 >= 8abc but the question suggests that a^4 + b^4 + c^4 + 16 = 8abc This is only possible when the am-gm inequality reduces to am-gm equality, and in am gm equality, we have all the terms to be equal to each other. thus we have, a=b=c=2 which gives abc=8

Using the inequality between aritmetic media and harmonic media, we have:

$\frac{ a^4 + b^4 + c^4 + 16}{4} \ge \sqrt[4]{a^4b^4c^416}$

Therefore, $a^4+b^4+c^4 \ge 8abc$ and the equality happen when $a^4=b^4=c^4=16$ . So, $a=b=c=2$ and $abc=8$

From Arithmetic-Geometric mean, for $a,b,c$ positive integers we have $a^4+b^4+c^4+2^4\ge 4\sqrt[4]{a^4.b^4.c^4.2^4}=8abc$ . Because we have $a^4+b^4+c^4+16=8abc$ , then equaity holds, which means that $a=b=c$ . we can conclude that $a,b,c,$ and $2$ are the roots from the polynomial $3x^4-8x^3+16=0$ , then $a.b.c.2=16$ , so $abc=8$ .

Since a,b,c are real positive numbers, A.M>= G.M inequality can be used on (a^4),(b^4),(c^4) and . on (2^4). In the given question the R.H.S is the G.M, so equality occurred, so a=b=c=2. So abc=8.

Since a,b,c are real positive, we can analyze AM>=GM: 8abc/4 >= [(2abc)^4]^1/4 =2abc. When the equality occurs means that all terms are equal: a=b=c=2. Therefore: abc=8.

Consider AM-GM inequality $\frac{a^4 + b^4 + c^4 + 2^4}{4} \geq \sqrt[4]{a^4 \cdot b^4 \cdot c^4 \cdot2^4}$ $a^4 + b^4 + c^4 + 16 \geq 8abc$

The equality is satisfied when $a=b=c=2$ Hence value of $abc=8$

By the AM-GM inequality, $a^4+b^4+c^4+16=a^4+b^4+c^4+(2)^4 \ge 4(a)(b)(c)(2)=8abc$ , and equality holds iff $a^4=b^4=c^4=16$ , i.e. $a=b=c=2$ . Hence, the value of $abc$ is $(2)(2)(2)=\boxed{8}$ .

$a^4$ + $b^4$ + $c^4$ + $2^4$ =4(2abc). or we can write

[ $a^4$ + $b^4$ + $c^4$ + $2^4$ ]/4≥( $a^4$ $b^4$ $c^4$ . $2^4$ ) $^[1/4]$

AM≥GM equality holds if a=b=c=2.

so abc=8

By AM-GM inequality we have $\frac{a^{4}+b^{4}+c^{4}+2^{4}}{4}\geq2abc$ with equality iff $a=b=c=2$ , hence $abc=8$ .

Observe that by AM-GM inequality, you will realize that the given condition is an equality condition. This means that a = b = c = 2. Hence abc = 8.

By $AM-GM$ inequality, $a^4+b^4+c^4+d^4\geq4abcd$ ,where equality holds at $a=b=c=d$ .Let $d=2$ yields the given equation to hold when $a=b=c=d=2$ .Then $abc=8$

Represent 16 as 2^4 and make 8abc expressed as 4(abc(2)). We have: (a^4 + b^4 + c^4 + 2^4)/4 = a * b * c * 2 This is the equality situation of the AM-GM on on a, b, c, and 2, which is only satisfied by having a=b=c=2, so abc = 8 .

Let us apply the $A.M.$ $\geq$ $G.M.$ inequality on the four positive reals $a^4$ , $b^4$ , $c^4$ and $16$ ,
Then we shall get,
$\frac{a^4+b^4+c^4+16}{4}$ $\geq$ $\sqrt[4]{16a^4b^4c^4}$
⇒ $a^4+b^4+c^4+16$ $\geq$ $4.2abc$
⇒ $a^4+b^4+c^4+16$ $\geq$ $8abc$
But we know that, $a^4+b^4+c^4+16$ $=$ $8abc$
We know that, in the $A.M.$ $\geq$ $G.M.$ inequality, the equality holds only when all the terms are equal.
So, $a^4$ $=$ $b^4$ $=$ $c^4$ $=$ $16$
⇒ $a$ $=$ $b$ $=$ $c$ $=$ $2$
So, $abc$ = $2.2.2$ = $8$ [ANSWER]

Since $a^4, b^4, c^4, 2^4$ are positive, we can apply AM-GM to $a^4, b^4, c^4, 2^4$ :

$\dfrac{a^4 + b^4 + c^4 + 2^4}{4} \ge \sqrt[4]{a^4 b^4 c^4 2^4}$

$\dfrac{a^4 + b^4 + c^4 + 16}{4} \ge 2abc$

$a^4 + b^4 + c^4 + 16 \ge 8abc$

Note that the problem achieves an equality, so we must have $a^4 = b^4 = c^4 = 16$ (AM-GM achieves an equality only when all its arguments are equal). Since $a,b,c$ are positive, we have $a = b = c = 2$ and so $abc = \boxed{8}$ .

$a^4+b^4+c^4+2^4$ = $4*(2abc)$ With AM-GM Equality, we can write it as: $(a^4+b^4+c^4+2^4)/4$ ≥ $(a^4*b^4*c^4*2^4)^{1/4}$

To make The AM-GM Equality holds we need to make sure that a=b=c.

The only number that can make it true is 2, so a=b=c=2.

Then, abc=2.2.2=8

If we use AM-GM on the left side we get that $\frac{a^4+b^4+c^4+2^4}{4} \ge 2abc$ or $a^4+b^4+c^4+2^4 \ge 8abc$ .

The equality case of this looks like the given equation and we know that equality is met in AM-GM only if all the terms are equal or $a^4=b^4=c^4=2^4$ which gives $a=b=c=2$ since $a,b,c$ are positive integers. Checking to see if this works we get $4 \times 16 = 8 \times 2^3$ which is true. Therefore $abc = \boxed{8}$ .

Notice that using the AM-GM inequality on the LHS yields the RHS. This means that a^4, b^4, c^4,, and 16 must be equal, meaning that a=b=c=2, and the answer is 8.

By AM-GM inequality, we have $a^4 + b^4 +c^4 + 2^4 \geq 4*2abcd=8abcd$ $$ We know that equality in AM-GM exists when all the terms are equal. This implies $a^4 = b^4 = c^4 =2^4$ which imples $a=b=c=2$ because $a,b,c$ are positive reals. $$ Hence $abc=2*2*2 = \boxed{8}$

We will use AM-GM inequality $(\frac{a^{4} + b^{4} +c^{4} + 16}{4}) \geq (16a^{4}b^{4}c^{4})^{1/4}$ we get AM =GM, $\Rightarrow$ a = b =c = 2 or abc =2

Let P = (a^{4} + b^{4} +c^{4} + 16) \geq 4 sqrt[4]{16 a^{4}b^{4}c^{4}} (Using AM-GM inequality) \Rightarrow P \geq 8 a b*c since P = 8abc, AM = GM \Rightarrow a = b = c = 2 \Rightarrow abc=2

Consider the $4$ th powers modulo $8$ , they comprise only $0$ & $1$ . $8$ divides the LHS, so $8 | (a^4 + b^4 + c^4)$ But none of 3-combinations of 0 & 1 can add up to $0 \pmod{2}$ , except if $8 | a^4$ , $8 | b^4$ , $8 | c^4$ $\Rightarrow 2 | a, b, c$ . Clearly, $a = b = c = 2$ suffices.

Note that the LHS must be even, since it has a factor of $8$ . Note that the LHS has the same parity as $a^4 + b^4 + c^4$ , which has the same parity as $a + b + c$ . Thus, $a + b + c$ must be even. Simply trying $a = b = c = 2$ works!

Specifically: $2^4 + 2^4 + 2^4 + 16 = 8 \cdot 2 \cdot 2 \cdot 2,$ so $abc = \boxed {8}$ .