# An inequality with $a,b,c>0$

Let $a,b,c>0$. Prove that: $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq\left(1+\sqrt{2}+\sqrt{3}\right)\cdot\left(\frac{1}{a+b\sqrt{2}+c\sqrt{3}}+\frac{1}{b+c\sqrt{2}+a\sqrt{3}}+\frac{1}{c+a\sqrt{2}+b\sqrt{3}}\right)$$

• The sign is reversed.
– user27126
Jan 15, 2014 at 2:50
• Where did you find this question - it is certainly wrong. For e.g. $(a, b, c) = (0.1, 0.1, 2.8)$ gives $LHS \approx 20.36 > RHS \approx 3.13$. Jan 15, 2014 at 3:46

It will suffice to show that the following (stronger) inequality holds for positive $a,b,c,u,v,w$ :
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq (u+v+w)\bigg(\frac{1}{au+bv+cw}+\frac{1}{av+bw+cu}+\frac{1}{aw+bu+cv}\bigg)$$
The fonction $f : x \mapsto \frac{1}{x}$ is convex on $(0,\infty)$, so
$$\begin{array}{lcl} f(\frac{u}{u+v+w}a+\frac{v}{u+v+w}b+\frac{w}{u+v+w}c) &\leq& \frac{u}{u+v+w}f(a)+\frac{v}{u+v+w}f(b)+\frac{w}{u+v+w}f(c) \\ f(\frac{v}{u+v+w}a+\frac{w}{u+v+w}b+\frac{u}{u+v+w}c) &\leq& \frac{v}{u+v+w}f(a)+\frac{w}{u+v+w}f(b)+\frac{u}{u+v+w}f(c) \\ f(\frac{w}{u+v+w}a+\frac{u}{u+v+w}b+\frac{v}{u+v+w}c) &\leq& \frac{w}{u+v+w}f(a)+\frac{u}{u+v+w}f(b)+\frac{v}{u+v+w}f(c) \\ \end{array}$$