×

# 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, 1 month ago

Sort by:

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,

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

Oh wow, that is a coincidence. I did exactly the same on both the proofs. · 2 years, 1 month 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. · 2 years, 1 month ago

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! · 2 years, 1 month ago

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? · 2 years, 1 month ago