Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

Question

Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$

I tried various methods. But, couldn't solve it. It'd be great if anyone can help.

That looks like AM-GM. Tried with a suitable combination of te ms? — Macavity, Apr 21 '16 at 04:01
@Macavity I tried everything I could come up with. Nothing has worked so far. — Sntn, Apr 21 '16 at 04:17
That's a substantial change to do after 6 hours, even though it can be tackled easily with AM-GM. Best to post a different question and reverse this change. — Macavity, Apr 21 '16 at 09:26

score 5 · Accepted Answer · edited Apr 21 '16 at 10:52

5

By AM-GM

$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \frac{2\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}} = \frac2a+\frac2b+\frac2c$$

For the last equality $abc=1$ was used.

Then again by AM-GM; $$\frac1a+\frac1b+\frac1c \ge \frac3{\sqrt[3]{abc}}=3$$

and again by a well known inequality $$\frac1a+\frac1b+\frac1c \ge \frac1{\sqrt{bc}}+\frac1{\sqrt{ca}}+\frac1{\sqrt{ab}}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$

For the last equality $abc=1$ was used. Combine the last two inequalities to get the desired result.

edited Apr 21 '16 at 10:52

gebruiker

6,272

answered Apr 21 '16 at 09:14

Sayan

2,728

Nice +1. Your well known inequality can be shown easily through AM-GM or Rearrangement... Sad thing is the OP not doing his share of basic effort. – Macavity Apr 21 '16 at 09:32

score 4 · Answer 2 · answered Apr 21 '16 at 06:13

4

You already have a proof using AM-GM and Muirhead. For a pure AM-GM proof, note that $\frac12 \left(\frac{a}b+\frac{b}a\right) \geqslant 1$ and so on, so half the LHS takes care of the $3$ on RHS.

For the rest, note you can cyclically sum the (weighted) AM-GMs: $$\frac5{18} \left(\frac{a}b+\frac{a}c\right)+\frac2{18} \left(\frac{b}a+\frac{b}c\right)+\frac2{18} \left(\frac{c}a+\frac{c}b\right)\geqslant \frac{\sqrt[3]a}{\sqrt[6]{b c}}=\sqrt{a}$$

answered Apr 21 '16 at 06:13

Macavity

48,039

This is somewhat more understandable to me. – Sntn Apr 21 '16 at 07:27
1

If it helps, you could do the second in two steps, $a/b+a/c \ge 2a/\sqrt{bc} = 2\sqrt{a^3}$ and then show a convex combination of $\sqrt{a^3}, \sqrt{b^3}, \sqrt{c^3}$ can you $\sqrt{a}$ etc. – Macavity Apr 21 '16 at 08:08
I've updated the question. It was a typing mistake. I apologize. Can you work it out now? – Sntn Apr 21 '16 at 08:59
1

That's too large a typo after too long. Not going to try if you keep changing the question. – Macavity Apr 21 '16 at 09:28
I know. But, this one is correct. – Sntn Apr 21 '16 at 10:19

Michael Rozenberg · Answer 3 · 2016-04-21T09:08:59.803

2

Id est, we need to prove that $\sum\limits_{cyc}(a^2b+a^2c-a^{\frac{4}{3}}b^{\frac{5}{6}}c^{\frac{5}{6}}-abc)\geq0$

which is AM-GM and Muirhead because $(2,1,0)\succ\left(\frac{4}{3},\frac{5}{6},\frac{5}{6}\right)$

Your another problem can be solved by the same way.

edited Apr 21 '16 at 09:08

answered Apr 21 '16 at 04:41

Michael Rozenberg

203,855

1

Sorry. But, I cannot understand what you said. Can you please explain a bit more? – Sntn Apr 21 '16 at 05:03
1

@SayantanSantra You might read a bit about Muirhead's inequality and AM-GM inequality on Wikipedia. (And in many other places.) – Martin Sleziak Apr 21 '16 at 07:25

Solumilkyu · Answer 4 · 2016-04-21T11:53:02.057

The original post was to show that $$\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3. $$ Observe that \begin{align} \frac{a}{b}+\frac{c}{a}&\ge2\sqrt{\frac{c}{b}}=2c\sqrt{a},\\ \frac{b}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c}{a}}=2c\sqrt{b},\\ \frac{c}{a}+\frac{c}{b}&\ge2\sqrt{\frac{c^2}{ab}}=2c\sqrt{c}. \end{align} and similarly \begin{align} &\frac{a}{c}+\frac{b}{a}\ge2b\sqrt{a}, &\frac{b}{a}+\frac{b}{c}\ge2b\sqrt{b}, &&\frac{c}{a}+\frac{b}{c}\ge2b\sqrt{c};\\ &\frac{a}{b}+\frac{a}{c}\ge2a\sqrt{a}, &\frac{b}{c}+\frac{a}{b}\ge2a\sqrt{b}, &&\frac{c}{b}+\frac{a}{c}\ge2a\sqrt{c}. \end{align} So, by combining them, and use the fact that $a+b+c\ge 3\sqrt[3]{abc}=3$, we have \begin{align} 3\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right) &\ge 2(a+b+c)\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\\ &\ge 6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right). \end{align} That is, $$\frac{1}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}.\tag{1}$$ Next, we also have \begin{align} \frac{1}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right) &= \frac{1}{2}\left[\left(\frac{b}{a}+\frac{a}{b}\right) +\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\right]\\ &\ge \frac{1}{2}(2+2+2)\\ &=3.\tag{2} \end{align} Hence the result follows by combining $(1)$ and $(2)$.

Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$

4 Answers4