一道中国高考原题,不服来做

Algebra Level 3

S = x + 2 + y + 5 + z + 10 T = x + 1 + y + 1 + z + 1 \begin{aligned} S&=\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}\\\\ T&=\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1} \end{aligned}

If the above is true for x , y , z > 0 , x, y, z >0, what is the minimum value of S 2 T 2 ? S^{2}-T^{2}?


The answer is 36.

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7 solutions

Yang Daniel
Sep 7, 2018

we let a = x + 1 , b = y + 1 , c = z + 1 a=\sqrt{x+1},b=\sqrt{y+1},c=\sqrt{z+1}

then T = a + b + c T=a+b+c ,and S = a 2 + 1 2 + b 2 + 2 2 + c 2 + 3 2 S=\sqrt{a^2+1^2}+\sqrt{b^2+2^2}+\sqrt{c^2+3^2}

now consider this graph

Then A E = a 2 + 1 2 , E F = ( b 2 + 2 2 ) , F G = ( c 2 + 3 2 ) AE=\sqrt{a^2+1^2},EF=\sqrt(b^2+2^2),FG=\sqrt(c^2+3^2)

So T = A B + B C + C D = A D , a n d , S = A E + E F + F G T = AB+BC+CD=AD, and,S = AE+EF+FG

And then

S 2 T 2 = ( A E + E F + F G ) 2 A D 2 A G 2 A D 2 = G D 2 = 36 S^2-T^2 = (AE+EF+FG)^2-AD^2 \ge AG^2-AD^2=GD^2=36

when A,F,G in the same line the inequation holds.

Beautiful!!

Phil Greene - 2 years, 8 months ago

Yeah I solved it with the obvious and quicker one but this way is really amazing.

Geeta . - 2 years, 8 months ago

I did the same as you. Love this solution!

X X - 2 years, 8 months ago

Excellent solution!

Krishnaraj Sambath - 2 years, 7 months ago
Kelvin Hong
Sep 16, 2018

I think Cauchy-Schwarz is way more straightforward:

S 2 T 2 = ( S + T ) ( S T ) = [ ( x + 2 x + 1 ) + ( y + 5 y + 1 ) + ( z + 10 z + 1 ) ] [ ( x + 2 + x + 1 ) + ( y + 5 + y + 1 ) + ( z + 10 + z + 1 ) ] = [ ( x + 2 x + 1 ) + ( y + 5 y + 1 ) + ( z + 10 z + 1 ) ] [ 1 x + 2 x + 1 + 4 y + 5 y + 1 + 9 z + 10 z + 1 ] ( 1 + 2 + 3 ) 2 = 36 \begin{aligned}S^2-T^2&=(S+T)(S-T)\\ &=\bigg[(\sqrt{x+2}-\sqrt{x+1})+(\sqrt{y+5}-\sqrt{y+1})+(\sqrt{z+10}-\sqrt{z+1})\bigg]\bigg[(\sqrt{x+2}+\sqrt{x+1})+(\sqrt{y+5}+\sqrt{y+1})+(\sqrt{z+10}+\sqrt{z+1})\bigg]\\ &=\bigg[(\sqrt{x+2}-\sqrt{x+1})+(\sqrt{y+5}-\sqrt{y+1})+(\sqrt{z+10}-\sqrt{z+1})\bigg]\bigg[\dfrac{1}{\sqrt{x+2}-\sqrt{x+1}}+\dfrac4{\sqrt{y+5}-\sqrt{y+1}}+\dfrac9{\sqrt{z+10}-\sqrt{z+1}}\bigg]\\&\geq (1+2+3)^2\\&=36\end{aligned}

The equality holds when

{ x + 2 x + 1 = 1 y + 5 y + 1 = 2 z + 10 z + 1 = 3 \begin{cases}\sqrt{x+2}-\sqrt{x+1}=1\\\sqrt{y+5}-\sqrt{y+1}=2\\\sqrt{z+10}-\sqrt{z+1}=3\end{cases} Which is accurately x = y = z = 0 x=y=z=0 .

but x=y=z=0 is not allowed, [math]x,y,z>0[/math]

Vilim Lendvaj - 2 years, 8 months ago

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(x,y,z)=(3,15,35) is allowed, too.

X X - 2 years, 8 months ago

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But it is not trivial right? How to find it?

Kelvin Hong - 2 years, 8 months ago

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@Kelvin Hong View Yang Daniel's solution, it clearly shows that the minimum occurs at x + 1 : y + 1 : z + 1 = 1 : 2 : 3 \sqrt{x+1}:\sqrt{y+1}:\sqrt{z+1}=1:2:3

X X - 2 years, 8 months ago

x = y = z = 0 x=y=z=0 is NOT a solution. In that case S 2 T 2 37.41096 S^2-T^2\approx 37.41096 . Perhaps you meant x = 0 , y = 3 , z = 8 x=0, y=3, z=8 ?

With regards to the x > 0 x>0 comment: if you take x x infinitesimally greater than 0, you will be able to get S 2 T 2 S^2-T^2 arbitrarily close to 36. Close enough to be able to convincingly state that the minimum value of 36 is indeed possible, even though it is reached at the border of the domain.

As X X mentions in response to the comment below infinitely many solutions for S 2 T 2 = 36 S^2-T^2=36 do exist, also completely inside the domain.

Roland van Vliembergen - 2 years, 8 months ago

sqrt(z+10) - sqrt(z+1) is not equal to 3 for any integere z.

Sqrt(15+10) - sqrt(15+1) = 1

sqrt(0+10) - sqrt(0+1) = sqrt(10) - 1 is less than 3.

Kermit Rose - 2 years, 4 months ago
Kirigaya Kazuto
Aug 30, 2018

By cauchy inequality S 2 T 2 = ( S + T ) ( S T ) = ( x + 2 + y + 5 + z + 10 + x + 1 + y + 1 + z + 1 ) ( x + 2 + y + 5 + z + 10 x + 1 y + 1 z + 1 ) = ( x + 2 + x + 1 + y + 5 + y + 1 + z + 10 + z + 1 ) ( 1 x + 2 + x + 1 + 4 y + 5 + y + 1 + 9 z + 10 + z + 1 ) ( 1 + 2 + 3 ) 2 = 36 \begin{aligned} S^{2}-T^{2} &=(S+T)(S-T)\\&=(\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}+\sqrt{x+1}+\sqrt{y+1}+\sqrt{z+1})(\sqrt{x+2}+\sqrt{y+5}+\sqrt{z+10}-\sqrt{x+1}-\sqrt{y+1}-\sqrt{z+1})\\&=(\sqrt{x+2}+\sqrt{x+1}+\sqrt{y+5}+\sqrt{y+1}+\sqrt{z+10}+\sqrt{z+1})(\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{4}{\sqrt{y+5}+\sqrt{y+1}}+\frac{9}{\sqrt{z+10}+\sqrt{z+1}})\\&≧(1+2+3)^{2}\\&=\boxed{36} \end{aligned}

Hi, there are typos at third last line.

Brian Lie - 2 years, 9 months ago

Sir, your reasoning is wrong, base on your calculation the equality required negative value of x , y , z x,y,z to hold.

Kelvin Hong - 2 years, 8 months ago

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(x,y,z)=(3,15,35) is allowed, too.

X X - 2 years, 8 months ago

thank you, the mistake has been corrected

Kirigaya Kazuto - 2 years, 8 months ago

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NO! Now the equality doesn't hold also.

Kelvin Hong - 2 years, 8 months ago

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Why the equality doesn't hold ?you know there is almost no difference between your method and mine

Kirigaya Kazuto - 2 years, 8 months ago

对x,y,z>0,取不到最小值36吧?要x=y=z=-1才取到36,请问原题出自哪呢?

Lean Boole - 2 years, 8 months ago

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(x,y,z)=(3,15,35) is allowed, too.

X X - 2 years, 8 months ago

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Wow, amazing! Thank you.

Lean Boole - 2 years, 8 months ago
K T
Sep 22, 2018

Since S 2 T 2 = ( S + T ) ( S T ) S^2-T^2=(S+T)(S-T) , the product rule gives us d ( S 2 T 2 ) d x = 2 S d S d x 2 T d T d x \frac{d(S^2-T^2)}{dx} = 2S\frac{dS}{dx}-2T\frac{dT}{dx} .

At minimum this must be 0, so S x + 2 = T x + 1 \frac{S}{\sqrt{x+2}} = \frac{T}{\sqrt{x+1}} S 2 T 2 = x + 2 x + 1 \frac{S^2}{T^2} = \frac{x+2}{x+1} S 2 T 2 1 = 1 x + 1 \frac{S^2}{T^2} -1 = \frac{1}{x+1} Doing this for y and z as well we get S 2 T 2 1 = 1 x + 1 = 4 y + 1 = 9 z + 1 \frac{S^2}{T^2} -1 = \frac{1}{x+1} = \frac{4}{y+1} = \frac{9}{z+1} Introducing a constant k such that 1 k = S 2 T 2 1 \frac{1}{k}=\frac{S^2}{T^2} -1 and using the original definition T = 1 + x + 1 + y + 1 + z T=\sqrt{1+x}+\sqrt{1+y}+\sqrt{1+z} , we can express T at minimum in terms of k: T = k + 4 k + 9 k = 6 k T=\sqrt{k}+\sqrt{4k}+\sqrt{9k}=6\sqrt{k} T 2 = 36 k T^2=36k

And using the definition for k, we find S 2 = ( 1 k + 1 ) T 2 = 36 + 36 k S^2=(\frac{1}{k}+1)T^2=36+36k In the difference k does not appear, so we have our answer: S 2 T 2 = 36 S^2-T^2=36

Albert Yiyi
Sep 3, 2018

Recall that by QM-AM inequality, for any positive a , b a,b , a 2 + b 2 2 a + b 2 \sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2} with equality if and only if a = b a=b .

let x + 1 = a , x + 2 = a 2 + 1 , a > 1 y + 1 = b , y + 5 = b 2 + 4 , b > 1 z + 1 = c , z + 10 = c 2 + 9 , c > 1 S = a 2 + 1 + b 2 + 4 + c 2 + 9 T = a + b + c S 2 T 2 = ( a 2 + 1 + b 2 + 4 + c 2 + 9 ) 2 T 2 = 2 ( a 2 + 1 2 2 + b 2 + 4 2 + c 2 + 9 2 ) 2 T 2 2 ( a + 1 2 + b + 2 2 + c + 3 2 ) 2 T 2 = ( a + b + c + 6 ) 2 2 ( a + b + c ) 2 \begin{aligned} \text{let } \sqrt{x+1} &= a , \sqrt{x+2} = \sqrt{a^2+1} , a>1 \\ \sqrt{y+1} &= b , \sqrt{y+5} = \sqrt{b^2+4} , b>1 \\ \sqrt{z+1} &= c , \sqrt{z+10} = \sqrt{c^2+9}, c>1 \\ \\ S &= \sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9} \\ T &= a+b+c \\ \\ S^2-T^2 &= (\sqrt{a^2+1}+\sqrt{b^2+4}+\sqrt{c^2+9})^2 - T^2 \\ &= 2 \left( \sqrt{\frac{a^2+1^2}{2}}+\sqrt{\frac{b^2+4}{2}}+\sqrt{\frac{c^2+9}{2}} \right)^2 - T^2 \\ &\geq 2 \left( \frac{a+1}{2}+\frac{b+2}{2}+\frac{c+3}{2} \right)^2 - T^2 \\ &= \frac{(a+b+c+6)^2}{2} - (a+b+c)^2 \end{aligned}

with equality if and only if a = 1 , b = 2 , c = 3 a=1, b=2, c=3 , min ( S 2 T 2 ) = ( 1 + 2 + 3 + 6 ) 2 2 ( 1 + 2 + 3 ) 2 = 36 \min(S^2-T^2) = \frac{(1+2+3+6)^2}{2} - (1+2+3)^2 = 36

note: there is a minor problem , to reach this minimum value, it must be a = 1 , x = 0 a=1, x=0 but x > 0 x>0 . so this minimum value is unreachable.

edit: ( x , y , z ) = ( 1 , 7 , 17 ) (x,y,z)=(1,7,17) would make it to 36 36 . (thanks to Michael Mendrin)

Try ( x , y , z ) = ( 1 , 7 , 17 ) (x,y,z)=(1,7,17)

( 3 + 12 + 27 ) 2 ( 2 + 8 + 18 ) 2 = 36 ( \sqrt{3}+\sqrt{12}+\sqrt{27})^2- (\sqrt{2}+\sqrt{8}+\sqrt{18})^2 = 36

Michael Mendrin - 2 years, 9 months ago

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good point. not sure where is the flaw in my reasoning though...

albert yiyi - 2 years, 9 months ago

personally i think it's a huge problem, and it doesn't solve the problem. So abandon this way will be better :(

晟轶 龚 - 2 years, 9 months ago
왕자 어린
Sep 18, 2018

-Thanks for Cauchy-Schwarz S^2-T^2=(S+T)(S-T), clearly

Vinod Kumar
Sep 17, 2018

Used Wolfram alpha to get the following four answers all giving 36 as the local minimum.

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.100514, 3.40206, 8.90463)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.724565, 5.89826, 14.5211)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.567207, 5.26883, 13.1049)

min{(sqrt(x + 2) + sqrt(y + 5) + sqrt(z + 10))^2 - (sqrt(x + 1) + sqrt(y + 1) + sqrt(z + 1))^2|x>0 ∧ y>0 ∧ z>0} = 36 at (x, y, z)≈(0.543924, 5.1757, 12.8953)

There might be are more such local minimum. I have printed four of them.

Answer=36

Use LaTeX; it is hard to read

Syed Hamza Khalid - 2 years, 8 months ago

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