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Let a,b,c be positive real numbers such that $abc=1$. Prove that $$\frac 1{a^3(b+c)}+\frac 1{b^3(c+a)}+\frac 1{c^3(a+b)} \ge \frac 32$$ We can derive the following inequalities from the given equality :-$$ab+bc+ca\ge3$$$$a+b+c\ge 3$$$$\frac 1a+\frac 1b+\frac 1c\ge3$$ None of them help to get a solution. If we simplify the given inequality, we get the following $$(a+b)(b+c)(\frac 2{b^3}-a-c)+(b+c)(c+a)(\frac 2{c^3}-a-b)+(c+a)(a+b)(\frac 2{a^3}-b-c) \ge 0$$ What do i do after this?

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3 Answers 3

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This is IMO 1995, Problem 2.

$\sum \frac{1}{a^3(b+c)}=\sum\frac{1/a^2}{a(b+c)}\ge\sum \frac{(1/a+1/b+1/c)^2}{2(ab+bc+ca)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{a^2b^2c^2}}{2}=\frac32.$

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  • $\begingroup$ please show us your own proof $\endgroup$ Dec 30, 2014 at 12:33
  • $\begingroup$ I think references to external answers are welcome, but should be posted as comments. Please, someone correct me if I'm wrong. $\endgroup$
    – fonini
    Jan 6, 2015 at 3:08
  • $\begingroup$ I like to post them as answers because that increases the chances that I will get reputation points. $\endgroup$ Jan 6, 2015 at 3:22
  • $\begingroup$ You might want to take a look here: meta.stackexchange.com/questions/225370/… $\endgroup$
    – fonini
    Jan 6, 2015 at 5:48
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$$x=\frac{1}{a}, y=\frac{1}{b} \text{ and } z=\frac{1}{c}$$

$$a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z} \text{ and } xyz=1.$$

$$\begin{align} LHS&=\frac {1}{a^3(b+c)}+\frac {1}{b^3(c+a)}+\frac {1}{c^3(a+b)}\\ &=\frac{1}{\frac{1}{x^3}\left (\frac{1}{y}+\frac{1}{z} \right)}+\frac{1}{\frac{1}{y^3}\left (\frac{1}{z}+\frac{1}{x} \right)}+...\\ &=\frac {x^3}{\frac {y+z}{yz}}+\frac {y^3}{\frac {z+x}{zx}}+\frac{z^3}{\frac{x+y}{xy}}\\ &=\frac{x^3yz}{y+z}+\frac{y^3zx}{z+x}+\frac{z^3xy}{x+y} \\ &=\frac{x^2xyz}{y+z}+\frac{y^2xyz}{z+x}+\frac{z^2xyz}{x+y}\\ &=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}.\end{align}$$

We will prove that $S\geq \frac{3}{2}$, where $S=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}$.

This inequality we multiply with $(b+c)+(c+a)+(a+b)$ and we obtain:

$$\begin{align}[(y+z)+(z+x)+(x+y)]\cdot S &=\left( \frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y} \right) [(y+z)+(z+x)+(x+y)]\\ &=x^2+y^2+z^2+ \left(x^2\frac{z+x}{y+z}+y^2\frac{y+z}{z+x} \right)+ \left(y^2\frac{x+y}{z+x}+z^2\frac{z+x}{x+y} \right)+...\\ &\geq x^2+y^2+z^2+2\sqrt{x^2y^2}+2\sqrt{y^2z^2}+2\sqrt{z^2x^2}\\ &=x^2+y^2+z^2+2xy+2yz+2zx\\ &=(x+y+z)^2.\end{align}$$

From here 2$(x+y+z)\cdot S\geq (x+y+z)^2,$ we have $S\geq \frac{x+y+z}{2}.$

Finally, $S\geq \frac{x+y+z}{2}=\frac{3}{2}\cdot \frac{x+y+z}{3}\geq \frac{3}{2}\cdot \sqrt[3]{xyz}=\frac{3}{2}\cdot \sqrt[3]{1}$

Equality holds for $x=y=z=1$, i.e. $a=b=c=1.$

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  • $\begingroup$ someone please edit the above answer...it is almost unreadable.. $\endgroup$ Dec 30, 2014 at 17:42
  • $\begingroup$ the proof is great but you could make it simpler after the transformation.. $\endgroup$ Dec 31, 2014 at 4:35
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This is my soenter image description herelution for this inequality

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  • $\begingroup$ Please typeset your answers using latex. Your answer might be deleted. $\endgroup$
    – wlad
    May 27, 2015 at 13:18

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