I'm working on some competition inequalities using some of the inequality theorems I just learned. I am trying to prove:

$$ \frac{a}{\sqrt{a^2 + 8bc}} + \frac{b}{\sqrt{b^2 + 8ca}} + \frac{c}{\sqrt{c^2 + 8ab}} \ge 1 $$

where $a$, $b$, and $c$ are positive reals. I recently learned about Jensen's inequality and am trying to apply it to this problem. I've tried something like this:

$$ \frac{a\sqrt{a}}{\sqrt{a^3 + 8abc}} + \frac{b\sqrt{b}}{\sqrt{b^3 + 8abc}} + \frac{c\sqrt{c}}{\sqrt{c^3 + 8abc}} \ge 1 $$

and am wondering if I can let:

$$ f(x) = \frac{x\sqrt{x}}{\sqrt{x^3 + 8p}} $$

(where $p = abc$) and use Jensen's inequality. Since $p$ depends on $a$, $b$, and $c$, am I not allowed to use this function for Jensen's inequality? Or because it's a "constant" can I use it anyway?

I've tried to test this for the sum:

$$ \frac{a}{bc} + \frac{b}{ca} + \frac{c}{ba} = \frac{a^2}{abc} + \frac{b^2}{abc} + \frac{c^2}{abc} $$

by defining the function $ f(x) = \frac{x^2}{p} $. Using Jensen's inequality gave me a correct result, but of course this isn't a proof that I can use this trick in general. Thoughts?


EDIT: I've gone ahead to apply the inequality anyway, and it turns out the $ f(x) $ I've defined for the problem is concave down, so it won't help in proving that the given sum is greater than or equal to another number. But it would still be nice to know for future reference if I could use this trick.

EDIT 2: Sorry about the wrong signs; it's fixed now!

  • 2
    $\begingroup$ As the inequality in question is homogeneous, you may in fact set $abc=1$ WLOG. Thus it's valid. $\endgroup$ – Macavity Aug 8 '15 at 0:22
  • $\begingroup$ In the denominators there should be a plus sign, not a minus one - artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/… $\endgroup$ – Jack D'Aurizio Aug 8 '15 at 3:06
  • $\begingroup$ Thanks for pointing out the wrong sign. $\endgroup$ – Guest Aug 8 '15 at 3:45

I remember how I solved it a long time ago. If we set $\frac{bc}{a^2}=x,\frac{ac}{b^2}=y,\frac{ab}{c^2}=z$, we have to prove: $$ \sum_{cyc}\frac{1}{\sqrt{1+8x}}\geq 1 $$ under the constraint $xyz=1$.

By setting $u=\sqrt{1+8x},v=\sqrt{1+8y},w=\sqrt{1+8z}$ we are left to prove: $$ \frac{1}{u}+\frac{1}{v}+\frac{1}{w}\geq 1 $$ under the constraints $u,v,w\geq 1$ and $(u^2-1)(v^2-1)(w^2-1)=512$.

Since the geometric mean is super-additive, we have: $$GM(u^2,v^2,w^2)\geq 1+GM(u^2-1,v^2-1,w^2-1) = 9 $$ and by the AM-GM inequality it follows that $u+v+w\geq 9$. With the same trick, $$ GM(u+1,v+1,w+1)\geq 1+GM(u,v,w) \geq 4,$$ so: $$(u+v+w-1)(u+1)(v+1)(w+1)\geq 8\cdot 4^3 = 512 $$ and by exploiting the constraint that is equivalent to: $$ u+v+w-1 \geq (u-1)(v-1)(w-1) $$ so: $$ uv+uw+vw \geq uvw $$ as wanted. For a solution using Jensen's inequality, see the link in the comments, or notice that the problem boils down to proving that $$ f(x)=\frac{1}{\sqrt{1+8x}}$$ is a convex function on $\mathbb{R}^+$. Pretty obvious since $f$ is positive and log-convex.

  • 1
    $\begingroup$ @Macavity: I think the OP put the wrong signs. It was a IMO problem. $\endgroup$ – Jack D'Aurizio Aug 8 '15 at 3:03
  • 1
    $\begingroup$ IMO problem 2001#2 : artofproblemsolving.com/wiki/index.php/2001_IMO_Problems/… $\endgroup$ – Jack D'Aurizio Aug 8 '15 at 3:05
  • 2
    $\begingroup$ Ah, there's even a solution by Jensen in that link. +1. $\endgroup$ – Macavity Aug 8 '15 at 3:06
  • 1
    $\begingroup$ Sorry about that! I copied down the wrong sign. Good answer!! $\endgroup$ – Guest Aug 8 '15 at 3:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.