Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


share|cite|improve this question
Welcome to math.SE. Since you are new, I want to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many users find the use of the imperative ("Find", "Show", etc) to be rude when asking for help. Please consider rewriting your post. – Arturo Magidin Jun 21 '12 at 19:30
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
up vote 6 down vote accepted

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 $\color{red}{2}$.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.