Guilherme's Inequalities - I

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

#Algebra #Inequality #GCDC

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 6 years, 3 months ago

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 - 6 years, 3 months ago

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 - 6 years, 3 months ago

@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 - 6 years, 3 months ago

@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 - 6 years, 3 months ago

@Prasun Biswas – Yes you are right. That is a typo.

Kartik Sharma - 6 years, 3 months ago

Oh wow, that is a coincidence. I did exactly the same on both the proofs.

Kartik Sharma - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$