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I need help with following inequality problem:

Let $a,b,c$ are positive real numbers $\mathbb{R}^+$. Prove that $$ \frac{2}{a^2}+\frac{5}{b^2}+\frac{45}{c^2}>\frac{16}{(a+b)^2}+\frac{24}{(b+c)^2}+\frac{48}{(a+c)^2}. $$ I was unable to find any classical inequality trick (AM-GM inequality, Cauchy-Schwarz…) which could work since the inequality is non-cyclic because of the numerators.

I tried to use trivial fact $\forall x,y \in \mathbb{R}^+$: $$ \frac{1}{x^2}+\frac{1}{y^2}>\frac{2}{(x+y)^2} $$ to create separate inequalities: $$ \frac{2}{a^2}+\frac{2}{b^2}>\frac{4}{(a+b)^2} $$ $$ \frac{3}{b^2}+\frac{3}{c^2}>\frac{6}{(b+c)^2} $$ but they not add up together to the desired form since term $\frac{42}{c^2}$ left on the left hand side and many others on the right hand side.

Clearing denominators (multiply everything) probably does not work since it leads to very complicated form.

I found out that numerators $24, 48$ are one less than perfect square ($24=5^2-1$, same for $48$) so I tried to use completing the square method but I did not succeed.

Oliver Bukovianský
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    You're really close. Here are some more observations to guide what you're thinking. 1/ Your trivial fact is way off. You'd want something like $\frac{1}{x^2} + \frac{1}{y^2} \geq \frac{8}{(x+y)^2}$. Hey, 8 on RHS! 2/ (Near-)Equality doesn't occur anywhere close to $k(1, 1, 1)$, whereas that's what your listed inequalities use. So those aren't what we really want. 3/ We'd likely want something like $ \frac{1}{x^2 } + \frac{ n}{y^2} \geq \frac{m } { (x+y)^2 } $. How do $n, m$ relate? When does equality hold? $\quad$ Combining with your existing ideas, it's possible to push this through. – Calvin Lin Aug 15 '24 at 20:17
  • @CalvinLin I conjectured that this inequality for $n \in \mathbb{N}, x,y \in \mathbb{R}^{+}$ holds: $\frac{1}{x^2}+\frac{n}{y^2}\geq \frac{3n+5}{(x+y)^2}$, but I do not know how to prove it after multiplying everything - I can not find nice factorization. – Oliver Bukovianský Aug 17 '24 at 07:21
  • After applying previous inequality for $n=14$ with $a,c$ and for $n=6$ with $b,c$ I get more promising inequality $(\frac{1}{a})^2+(\frac{2}{b})^2+(\frac{5}{c})^2>(\frac{4}{a+b})^2+(\frac{1}{b+c})^2+(\frac{1}{a+c})^2$ since only perfect squares are involved but I do not know how to prove it as well. – Oliver Bukovianský Aug 17 '24 at 12:02
  • 1/ I verified that your inequality is correct, likewise by expansion . 2/ Given that, you do have the desired inequality. Can you show me how you were combining them? I do not get your LHS nor RHS. – Calvin Lin Aug 17 '24 at 14:51

1 Answers1

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Prove the following. If you're stuck, explain what you've tried.

  1. I encourage you to guess what the 3 separate inequalities that we want are, before reading further. The RHS is pretty determined, so it's a matter of how to combine the coefficients on the LHS.

  2. $\frac{1}{x^2} + \frac{1}{y^2} \geq \frac{8}{(x+y)^2} $.

    • In particular, find a proof via one-application of the power mean inequality.
    • Equality holds when $x=y$.

  3. $\frac{1}{x^2} + \frac{n^2}{y^2} \geq \frac{8n}{(x+y)^2} $.

    • Equality holds only when $n = \frac{y}{x} = 1$.
    • My proof is essentially by expanding all terms, smartly.

  4. Using 3 inequalities for a suitable $n$ (see the RHS), arrive at the desired conclusion.

    • We have strict inequality as we use $n \neq 1$.
    • It is quite a weak bound, so there are likely other approaches. EG OP claims that $\frac{1}{x^2} + \frac{4}{y^2} \geq \frac{ 17}{(x+y)^2}$ which I've verified, whereas I'm only using $\frac{1}{x^2} + \frac{4}{y^2} \geq \frac{ 16}{(x+y)^2}$
Calvin Lin
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