Victor's Level 5 Challenges - 3

Algebra Level 5

There exists a constant C C such that

∑ i = 1 4 ( x i + 1 x i ) 3 ≥ C \sum_{i=1}^4\left(x_i+\frac{1}{x_i}\right)^3 \geq C

for all positive real numbers x 1 , x 2 , x 3 , x 4 x_1,x_2,x_3,x_4 which satisfy

x 1 3 + x 3 3 + 3 x 1 x 3 = x 2 + x 4 = 1. x_1^3+x_3^3+3x_1x_3=x_2+x_4=1.

Given that the largest constant C C can be expressed in the form m n \frac{m}{n} , where m m and n n are coprime, positive integers, find the value of m + n m+n .


The answer is 127.

Victor Loh by Victor Loh , 19, Singapore

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1 solution

From the equation, we get x 1 3 + x 3 3 + ( − 1 ) 3 = 3 x 1 x 2 ( − 1 ) x_{1}^{3}+x_{3}^{3} + (-1)^{3} = 3x_{1}x_{2}(-1)

Which gives x 1 + x 3 + ( − 1 ) = 0 x_{1}+x_{3}+(-1) = 0 or x 1 = x 3 = − 1 x_{1}=x_{3}=-1 .

Now we have x 1 + x 3 = 1 x_{1}+x_{3} = 1 , x 2 + x 4 = 1 x_{2}+x_{4} = 1

WLOG that x 1 = x 2 , x 3 = x 4 x_{1} = x_{2}, x_{3} = x_{4} to find the minimum.

I'll prove later that minimum occurs iff x 1 = x 3 = 1 2 x_{1}=x_{3}=\frac{1}{2} .

Therefore, ∑ i = 1 4 ( x i + 1 x i ) 3 ≥ 125 2 \displaystyle \sum\limits_{i=1}^{4} \left(x_{i}+\frac{1}{x_{i}}\right)^{3} \geq \boxed{\displaystyle \frac{125}{2}} ~~~

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Just following up with the last part

For convenience, let f ( x ) = x + 1 x f (x)=x+\frac {1}{x}

By power mean inequality, ( ∑ i = 1 4 f ( x i ) 3 4 ) 1 / 3 ≥ ∑ i = 1 4 f ( x i ) 4 (\frac {\sum_{i=1}^4 f (x_{i})^{3}}{4})^{1/3}\geq\frac {\sum_{i=1}^4 f (x_{i})}{4}

We wish to find the minimum value of the right hand side.

But x 1 + x 2 + x 3 + x 4 = 2 x_{1}+x_{2}+x_{3}+x_{4}=2 and by Cauchy-Schwarz inequality (or more directly Titu's lemma) ∑ i = 1 4 1 x i ≥ 16 x 1 + x 2 + x 3 + x 4 = 8 \sum_{i=1}^4 \frac {1}{x_{i}}\geq\frac {16}{x_{1}+x_{2}+x_{3}+x_{4}}=8 . The result follows.

Sorry, the previous post had formatting problems. How do I delete it?

Joel Tan - 6 years, 9 months ago

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On the top right there's a pencil icon there. Just click and delete it. ^_^

Samuraiwarm Tsunayoshi - 6 years, 9 months ago

I've deleted it for you :)

Victor Loh - 6 years, 9 months ago

Nice one! Too bad for me that I rarely use this inequality. XD

Samuraiwarm Tsunayoshi - 6 years, 9 months ago

To deal with the last part, it is slightly easier to simply apply Jensens to f ( x ) = ( x + 1 x ) 3 f(x) = ( x + \frac{1}{x} ) ^3 (check 2nd derivative) , which shows that the minimum of f ( x 1 ) + f ( x 3 ) f(x_1) + f(x_3) is achieved when x 1 = x 3 x_1 = x_3 .

Calvin Lin Staff - 6 years, 5 months ago

damn it, i don't realize this stuff of a^3+b^3+c^3=3abc only if a+b+c=0, I have a longer solution with Vie'te' for x1 and x3...

Nguyễn Phát - 6 years, 8 months ago

Can you explain how x 1 + x 3 + ( − 1 ) x_1 + x_3 + (-1) gives x 1 = x 3 = − 1 x_1 = x_3 = -1 ? I think you mean a + + instead?

Can you explain why "WLOG that x 1 = x 2 , x 3 = x 4 x_1 = x_2, x_3 = x_4 "?

Calvin Lin Staff - 6 years, 5 months ago

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