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# Inequality Problem

Can any one could help me Solve this problem ?

NOTE: This problem has been taken from KV JMO 2014

Note by Sudipta Biswas
2 years, 8 months ago

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Sorry for bumping this, but I saw that this remained unsolved, so I decided to have a go. From the other comments, it seems like there is a quick solution (given that it is a KV JMO problem, I did not expect it to be so long.) Here we go! (with some help from Wolfram Alpha in Method 2 - but the idea for the function was original, I promise.)

$$\text{Split the LHS as,} \\ \sqrt{a + b + c}\left(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}\right) + \dfrac{\sqrt{c}}{a + b} + \dfrac{\sqrt{b}}{c + a} + \dfrac{\sqrt{a}}{b + c} \\ \text{We have, by AM-HM or Cauchy-Schwarz (Titu's Lemma), } \\ \sqrt{a + b + c}\left(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}\right) \geq \sqrt{a + b + c} \cdot \dfrac{9}{2(a + b + c)} = \dfrac{9}{2\sqrt{a + b + c}} \\~\\ \text{The RHS begins to emerge! So we need to show, } \dfrac{\sqrt{c}}{a + b} + \dfrac{\sqrt{b}}{c + a} + \dfrac{\sqrt{a}}{b + c} \geq \dfrac{3\sqrt{3}}{2\sqrt{a + b + c}} \\~\\ \text{There are two ways I've found to prove this inequality. The first uses normalization and calculus,}\\ \text{while the other requires patience to factorize. I'm sorry.} \\~\\ \text{Method 1: The given inequality is homogeneous, so we can normalize as,} \\ a + b + c = 1 \Rightarrow a,b,c \in (0, 1) \\ \text{Now we need to show,} \\ \dfrac{\sqrt{a}}{1 - a} + \dfrac{\sqrt{b}}{1 - b} + \dfrac{\sqrt{c}}{1 - c} \geq \dfrac{3\sqrt{3}}{2}\\ \text{Let's focus on one of these terms. If we show, } \dfrac{\sqrt{a}}{1 - a} \geq \dfrac{3a\sqrt{3}}{2} \\ \text{Then, on addition of the three terms, we get the required inequality. On squaring, this is equivalent to,} \\ 4a \geq 27a^2(1 - a)^2 \Leftrightarrow 27a^3 - 54a^2 +27a - 4 \geq 0 \\~\\ \text{Consider the function, } f(x) = 27x^3 - 54x^2 + 27x - 4 \\ \text{On evaluating, we see } f(\frac{1}{3}) = f'(\frac{1}{3}) = 0 \\ \text{Also, } f(0) < 0, f(1) < 0 \\ \text{Thus, } f(x) \leq 0 \text{ } \forall \text{ } x \in (0, 1) \\ \text{But this is what we had to show for a, b and c!} \\~\\ \therefore \dfrac{\sqrt{a}}{1 - a} + \dfrac{\sqrt{b}}{1 - b} + \dfrac{\sqrt{c}}{1 - c} \geq \dfrac{3\sqrt{3}}{2}\\ \Rightarrow \dfrac{\sqrt{c}}{a + b} + \dfrac{\sqrt{b}}{c + a} + \dfrac{\sqrt{a}}{b + c} \geq \dfrac{3\sqrt{3}}{2\sqrt{a + b + c}} \\ \Rightarrow \sqrt{a + b + c}\left(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}\right) + \dfrac{\sqrt{c}}{a + b} + \dfrac{\sqrt{b}}{c + a} + \dfrac{\sqrt{a}}{b + c} \geq \dfrac{9 + 3\sqrt{3}}{2\sqrt{a + b + c}} \\ \text{And we are done.} \\~\\ \text{Method 2: to prove the inequality I was talking about, we show, } \\ \dfrac{\sqrt{a}}{b + c} \geq \dfrac{3a\sqrt{3}}{2(a + b + c)^{\frac{3}{2}}} \\ \text{This is equivalent to,} \\ (2a - b - c)^2 (a + 4 (b + c)) \geq 0 \\ \text{So we are done, again - on adding similar terms. Equality holds if, } a = b = c.$$ · 1 year, 4 months ago

Noticed that the expression can be written as

$\frac{1}{\sqrt{a+b+c}-\sqrt a}+\frac{1}{\sqrt{a+b+c}-\sqrt b}+\frac{1}{\sqrt{a+b+c}-\sqrt c}$

Then, by Titu's Lemma, we can combine everything into one term, but I don't know how to approach next. Anyone may continue? · 2 years, 8 months ago