Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2\ge \frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{12}{a^2+b^2+c^2}+\frac{18}{a^2+b^2+c^2+d^2}$

Question

Let $a,b,c,d$ be positive numbers. Show that $$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{12}{a^2+b^2+c^2}+\dfrac{18}{a^2+b^2+c^2+d^2}$$

I have seen this Similar Problem $$\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)^2\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2}$$

This problem pf: \begin{align} LHS &=\sum\dfrac{1}{a^2}+\sum\dfrac{2}{ab}\\ &=\dfrac{1}{a^2}+\dfrac{2}{ab}+(\dfrac{1}{b^2}+\dfrac{2}{ac}+(\dfrac{2}{ad}+\dfrac{2}{bc})++(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\ &\ge \dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+(\dfrac{1}{b^2}+\dfrac{4}{a^2+c^2})+(\dfrac{4}{a^2+d^2}+\dfrac{4}{b^2+c^2})+(\dfrac{1}{c^2}+\dfrac{1}{d^2}+\dfrac{2}{ab}+\dfrac{2}{bd}+\dfrac{2}{cd})\\ &\ge\dfrac{1}{a^2}+\dfrac{4}{a^2+b^2}+\dfrac{9}{a^2+b^2+c^2}+\dfrac{16}{a^2+b^2+c^2+d^2} \end{align}

This Problem is from PhamKim Hung-Secrets in Inequalities,P:25,can see:http://www.doc88.com/p-0919919061471.html — math110, Jun 28 '13 at 07:45
Pham Kim Hung’s proof shows your stronger inequality, so why do you ask the question ? Are you asking for a different proof ? — Ewan Delanoy, Jun 28 '13 at 08:09
That only poof other easy inequality.But this stronger inequality not have solution — math110, Jun 28 '13 at 08:15

score 1 · Accepted Answer · answered Jun 28 '13 at 08:26

There is the solution in the document you provided indeed - the proof is exactly the same. Note that on the top of p.25, there are four inequalities. Replace the third one by $$\frac{1}{b^2} + \frac{1}{c^2} \ge \frac{4}{b^2+c^2} \ge \frac{4}{a^2+b^2+c^2}$$ The fourth one by $$\frac{18}{ad+bd+cd} \ge \frac{18}{a^2+b^2+c^2+d^2}$$ and you get the result.

Prove that $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^2\ge \frac{1}{a^2}+\frac{4}{a^2+b^2}+\frac{12}{a^2+b^2+c^2}+\frac{18}{a^2+b^2+c^2+d^2}$

1 Answers1