Olympiad inequality problem with $a+b+c+abc=4$

Question

If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that

$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$ This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn't solve it. Can anyone help me? Thanks in advance:).

Answer 1

This inequality can be written like $$(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}})^2(a(b+c)+b(c+a)+c(a+b))\geq(a+b+c)^3$$ and this is true with Hölder's inequality

Answer 2

Consider using the following Hölder's inequality:

If $a_1,\ldots, a_n$ and $b_1,\ldots, b_n$ are positive real numbers, then $$\sum_{k = 1}^n a_kb_k \le \left(\sum_{k = 1}^n a_k^p\right)^{1/p}\left(\sum_{k = 1}^n b_k^q\right)^{1/q}$$ where $p$ and $q$ are real numbers such that $p,q \ge 1$ and $\frac{1}{p} + \frac{1}{q} = 1$.

Write

$$ab + bc + ca = \frac{ab + bc}{2}+\frac{bc + ca}{2} + \frac{ca + ab}{2} = \frac{a(b + c)}{2}+\frac{b(c + a)}{2}+\frac{c(a + b)}{2}.$$

Then

\begin{align}&\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3}\left(ab + bc + ca\right)^{1/3}\\ &=\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3} \left[\frac{a(b + c)}{2} + \frac{b(c + a)}{2} + \frac{c(a + b)}{2}\right]^{1/3}\\ &\ge \frac{a^{2/3}}{(b+c)^{1/3}}\cdot \frac{a^{1/3}(b+c)^{1/3}}{2^{1/3}} + \frac{b^{2/3}}{(c + a)^{1/3}}\cdot \frac{b^{1/3}(c+a)^{1/3}}{2^{1/3}} + \frac{c^{2/3}}{(a + b)^{1/3}}\cdot \frac{c^{1/3}(a + b)^{1/3}}{2^{1/3}}\\ &= \frac{a+b+c}{2^{1/3}}. \end{align}

The second to last step follows from Hölder's inequality with $p=3/2$ and $q=3$. Since $a + b + c = 4 - abc$, then

$$\left(\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c + a}} + \frac{c}{\sqrt{a+b}}\right)^{2/3}\left(ab + bc + ca\right)^{1/3} \ge \frac{4-abc}{2^{1/3}}.$$

The result is obtained by cubing both sides of this inequality.

Answer 3

Here is a possible approach without using Hölder's inequality. Assume without loss of generality that $a\ge b \ge c$. Then $$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2\ge \frac{1}{2a}\left(a+b+c\right)^2$$ Now $$ab+bc+ca=A=\frac{A+A}{2}=\frac{1}{2}(a(b+c)+b(c+a)+c(a+b))\ge c(a+b+c)$$ Since $a,b,c\in \mathbb{R}_+$ this implies $$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2(ab+bc+ca) \ge \frac{c}{2a}(a+b+c)^3$$ However $0<\frac{c}{a}\le 1$ by assumption, so the desired inequality follows.