0
$\begingroup$

I am stuck with the following problem :

If $a,b,c$ are positive and not all equal then prove that $$\,\,(a+\frac1a)^2+(b+\frac1b)^2+(c+\frac1c)^2 \geq 33\frac13,$$ when $a+b+c=1$.

Try: $\,\,(a+\frac1a)^2+(b+\frac1b)^2+(c+\frac1c)^2 \implies a^2+b^2+c^2+\dfrac{1}{ a^2}+\dfrac{1}{ b^2}+\dfrac{1}{ c^2}+6$ and then I did few more steps but could not come up with the result.

I will be grateful if someone explains it. Thanks in advance for your time.

$\endgroup$

marked as duplicate by Martin R, Lord Shark the Unknown, Community Aug 3 '17 at 6:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

You can you Jensen for $f(x)=x^2+\frac{1}{x^2}$ because $f''(x)=2+\frac{6}{x^4}>0$.

Hence, $$\sum_{cyc}\left(a^2+\frac{1}{a^2}\right)\geq3\left(\left(\frac{a+b+c}{3}\right)^2+\frac{1}{\left(\frac{a+b+c}{3}\right)^2}\right)=\frac{82}{3}.$$ Thus, $$\sum_{cyc}\left(a+\frac{1}{a}\right)^2\geq\frac{82}{3}+6=33\frac{1}{3}.$$

Also you can use C-S and Holder: $$\sum_{cyc}a^2+\sum_{cyc}\frac{1}{a^2}+6=\frac{1}{3}\sum_{cyc}1^2\sum_{cyc}a^2+\left(\sum_{cyc}a\right)^2\sum_{cyc}\frac{1}{a^2}+6\geq$$ $$\geq\frac{1}{3}(a+b+c)^2+(1+1+1)^3+6=33\frac{1}{3}.$$

Also, we can make the following thing.

We need to prove that $$\sum_{cyc}\left(a+\frac{1}{a}\right)^2\geq33\frac{1}{3}$$ or $$\sum_{cyc}\left(\left(a+\frac{1}{a}\right)^2-\frac{100}{9}\right)\geq0$$ or $$\sum_{cyc}\frac{(3a-1)(a-3)(3a^2+10a+3)}{a^2}\geq0$$ or $$\sum_{cyc}\left(\frac{(3a-1)(a-3)(3a^2+10a+3)}{a^2}+160(3a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(3a-1)^2(a^2+54a+9)}{a^2}\geq0.$$ Done!

$\endgroup$

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