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Guilherme's Inequalities - I

Let \( x,y,z \) be positive reals such that \( x+y+z=1 \). Show that \[ 24 < \dfrac{31x}{\sqrt{1+y+z}} + \dfrac{31y}{\sqrt{1+x+z}} + \dfrac{31z}{\sqrt{1+x+y}} < 32. \]

Note by Guilherme Dela Corte
2 years, 3 months ago

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I'm not sure if this is correct or not, but here it goes. I'll proceed to show that the lower bound of the inequality works but I still can't think of a proof for the upper bound.


Let \(f(x)=\dfrac{x}{\sqrt{2-x}}\). Plotting a graph of \(f(x)\) verifies that \(f(x)\) is a convex function. Take the given expression in the inequality as \(X\). Using the given data, we can write \(X\) as,

\[X=31\times \left(f(x)+f(y)+f(z) \right)\]

Applying Jensen's Inequality on \(f(x)\), we have,

\[\frac{f(x)+f(y)+f(z)}{3}\geq f\left(\frac{x+y+z}{3}\right)\\ \implies \frac{X}{93}\geq f\left( \frac{1}{3} \right)\\ \implies X\geq 93\times f\left( \frac{1}{3} \right)=93\times \frac{\frac{1}{3}}{\sqrt{2-\frac{1}{3}}}=31\times \sqrt{0.6}\approx 24.0125\\ \implies 24\lt X\\ \implies 24\lt \dfrac{31x}{\sqrt{1+y+z}} + \dfrac{31y}{\sqrt{1+x+z}} + \dfrac{31z}{\sqrt{1+x+y}}\] Prasun Biswas · 2 years, 3 months ago

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@Prasun Biswas Almost perfect approach on the lower bound - try a full proof on the convexity of the function through some derivatives. And I thought the upper bound would be easier to work on!

Please see that for \( 0 \leq x \leq 1 \), \(f(x) = \dfrac{x}{\sqrt{2-x}} \leq x \), and thus \( f(x) + f(y) + f(z) \leq x + y + z = 1 < \frac{32}{31} \) (weakening the bounding confirms the inequality). Guilherme Dela Corte · 2 years, 3 months ago

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@Guilherme Dela Corte Oh wow, that is a coincidence. I did exactly the same on both the proofs. Kartik Sharma · 2 years, 3 months ago

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@Guilherme Dela Corte Damn! Why didn't I think of that earlier?! Anyway, nice approach for the upper bound!

Another thing is that I still don't know how to prove the convexity of a function using calculus (the graph method is the only method I know of). And I couldn't find a good wiki on the topic. Prasun Biswas · 2 years, 3 months ago

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@Prasun Biswas Take the double derivative of the function and check whether \( f''(x) \geq 0 \) or \( f''(x) \leq 0 \) , for all \(x \in \text{Domain} \).

EDIT: Typo fixed, sorry! Guilherme Dela Corte · 2 years, 3 months ago

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@Guilherme Dela Corte Ah, thanks a bunch, man! By the way, I think you made a typo because it doesn't give the right result for, say \(f(x)=x^2\) which is strictly convex. Shouldn't the test be " \(f''(x)\leq 0\) (concave) or \(f''(x)\geq 0\) (convex), \(\forall x\in Dom(f)\) " ?

And the equality holds iff \(f(x)\) is not strictly convex or concave, I suppose? Prasun Biswas · 2 years, 3 months ago

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@Prasun Biswas Yes you are right. That is a typo. Kartik Sharma · 2 years, 3 months ago

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