# prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Assume: $a,b,c >0$ prove that :

$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

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How is $x,y,z$ related to $a,b,c$? – user17762 Jun 21 '12 at 19:30
Thank you Arturo Magidin,i found other problems were posted like this so i thought it is the right way. Marvis, i edited that . – Frank Jun 21 '12 at 19:51
Can we use Tchebyshev's inequality here? – Frank Jun 26 '12 at 11:32

Replace $(a,b,c)$ by $(x_1,x_2,x_3)$ for notational convenience, and start with the inequality $$\sum_{i,j}(x_i-x_j)\cdot\left(\frac1{x_j^2}-\frac1{x_i^2}\right)\geqslant0,$$ which holds because every term in the sum is nonnegative. Expanding the LHS, one gets $$\sum_{i,j}\frac{x_i}{x_j^2}\geqslant\sum_{i,j}\frac1{x_i}=3\cdot\sum_i\frac1{x_i}.$$ Separating the terms such that $i=j$ from those such that $i\ne j$ in the LHS yields $$\sum_{i\ne j}\frac{x_i}{x_j^2}\geqslant\color{red}{2}\cdot\sum_i\frac1{x_i},$$ which is strictly stronger than the inequality to prove thanks to the factor $\color{red}{2}$ in the RHS. Furthermore, this inequality is strict except when all the $x_i$s are equal. Finally, the same proof works for $n$ terms instead of $3$, yielding a factor $n-1$ instead of the factor $\color{red}{2}$.

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I assume that by $x,y,z$ you mean $a,b,c$.

Without loss of generality, we can assume that $c$ is the smallest of the three i.e. $a, b \geq c >0$. Then \begin{align} \dfrac{b+c}{a^2} + \dfrac{c+a}{b^2} + \dfrac{a+b}{c^2} - \dfrac1a - \dfrac1b - \dfrac1c & = \dfrac{b+c-a}{a^2} + \dfrac{c+a-b}{b^2} + \dfrac{a+b-c}{c^2}\\ & = \dfrac{b^3 +b^2c - ab^2 + a^2c + a^3 - a^2b}{a^2b^2} + \dfrac{a+b-c}{c^2}\\ & = \dfrac{c(a^2+b^2)+a^3+b^3-a^2b - ab^2}{a^2b^2} + \dfrac{a+b-c}{c^2}\\ & = \dfrac{c(a^2+b^2)+(a+b)(a^2-ab+b^2)-ab(a+b)}{a^2b^2} + \dfrac{a+b-c}{c^2}\\ & = \dfrac{c(a^2+b^2)+(a+b)(a^2-2ab+b^2)}{a^2b^2} + \dfrac{a+b-c}{c^2}\\ & = \dfrac{c(a^2+b^2)}{a^2b^2}+\dfrac{(a+b)(a-b)^2}{a^2b^2} + \dfrac{a+b-c}{c^2}\\ \end{align} Note that each term on the right side is non-negative. In fact, the first term (since $a,b,c >0)$ and last term (since $a,b \geq c > 0 \implies a+b > c$) are strictly positive. Hence, $$\dfrac{c(a^2+b^2)+(a+b)(a-b)^2}{a^2b^2} + \dfrac{a+b-c}{c^2} > 0$$ Hence, $$\dfrac{b+c}{a^2} + \dfrac{c+a}{b^2} + \dfrac{a+b}{c^2} - \dfrac1a - \dfrac1b - \dfrac1c > 0$$which gives us more than what we want.

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Using the AM-GM inequality we obtain: $$\frac{b+c}{a^{2}}+\frac{c+a}{b^{2}}+\frac{a+b}{c^{2}}=\frac{b^{3}c^{2}+b^{2}c^{3}+a^{2}c^{3}+a^{3}c^{2}+a^{3}b^{2}+a^{2}b^{3}}{a^{2}b^{2}c^{2}}\ge2\frac{a^{3}bc+b^{3}ac+c^{3}ab}{a^{2}b^{2}c^{2}}=2\frac{a^{2}+b^{2}+c^{2}}{abc}$$ Now a bit of juggling around proves an even stronger result: $$2\frac{a^{2}+b^{2}+c^{2}}{abc}=\frac{(a-b)^{2}+(a-c)^{2}+(b-c)^{2}+2(ab+bc+ac)}{abc}\ge2\frac{ab+bc+ac}{abc}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ In both estimates above equality is attained when $a=b=c$ which can be checked directly.

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HINT: One may start proving the inequality:

$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\ \frac{9}{a+b+c}+ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

that is easy to prove. I met some time ago this inequality and have just remembered now. Of course, it's easy only if you met it before, otherwise it's rather hard to make such a guess.

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