Ex-1.44(new book) (BOMC-2011)

[sm.joty]

If $a,b,c >0$ ,Prove that,
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{9}{a+b+c}$


It's very easy to get a relation between $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ and $\frac{9}{a+b+c}$ but how can we find any relation for the middle term.

[sourav das]

Hint

H.M.$\leq$A.M.

[Nadim Ul Abrar]

$(\frac{1}{a} +\frac{1}{b})/2 \geq \frac{1}{(ab)^{1/2}}$

$\frac{1}{ab^{1/2}} \geq \frac{2}{(a+b)}$ *

$\frac{1}{bc^{1/2}} \geq \frac{2}{(b+c)}$ *

$\frac{1}{ca^{1/2}} \geq \frac{2}{(c+a)}$ *
adding three stars we have
$\frac{1}{ab^{1/2}}+\frac{1}{bc^{1/2}}+\frac{1}{ca^{1/2}} \geq \frac{2}{(a+b)}+\frac{2}{(b+c)}+\frac{2}{(c+a)}$

note that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{1}{ab^{1/2}}+\frac{1}{bc^{1/2}}+\frac{1}{ca^{1/2}}$ ... We've proved first part ....

[Nadim Ul Abrar]

Second part

$\frac{(a+b)+(b+c)+(c+a)}{3} \geq \frac{3}{{\frac{1}{a+b}}+{\frac{1}{b+c}}+{\frac{1}{c+a}}}$

so proved