How to prove this inequality?

Posted August 23rd, 2007 by TommyR

For $a_1,a_2,a_3,a_4,a_5,a_6>0$ prove that:

$\frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+\frac{a_3}{a_4+a_5}+<br /> \frac{a_4}{a_5+a_6}+\frac{a_5}{a_6+a_1}+\frac{a_6}{a_1+a_2}\geq3$

Comments

On August 25th, 2007 Structure says:

As $f(x)=\frac{1}{x}$ is concave on $(0,\infty)$ :

$\sum_{i=1}^6\frac{\lambda_i}{x_i}\geq \frac{1}{\sum_{i=1}^{6}\lambda_ix_i}$
With $\sum_{i=1}^{6}\lambda_i = 1$ and $\lambda_i\geq0$

For $\lambda_i=\frac{a_i}{\sum_{i=1}^{6}a_i}$ and $x_i=a_{i+1}+a_{i+2}$ ( $a_7=a_1, a_8=a_2$ ) we have:

$\sum_{i=1}^6\frac{\frac{a_i}{\sum_{j=1}^6a_j}}{a_{i+1}+a_{i+2}}\geq<br /> \frac{1}{\sum_{i=1}^{6}{\frac{a_i}{\sum_{j=1}^{6}a_j}(a_{i+1}+a_{i+2})}}$

$\sum_{i=1}^6\frac{a_i}{a_{i+1}+a_{i+2}}\geq<br /> \frac{{(\sum_{j=1}^{6}a_j)}^2}{\sum_{i=1}^{6}{a_i}(a_{i+1}+a_{i+2})}$

We shall prove that:
$\frac{{(\sum_{j=1}^{6}a_j)}^2}{\sum_{i=1}^{6}{a_i}(a_{i+1}+a_{i+2})}\geq3$

Which is equivalent with:

${(a_1+a_4)}^2+{(a_2+a_5)}^2+{(a_3+a_6)}^2 \geq$
$\geq(a_1+a_4)(a_2+a_5)+(a_2+a_5)(a_3+a_6)+(a_3+a_6)(a_1+a_4)$

This is true because:
$a^2+b^2+c^2\geq ab+bc+ca$