Another inequality

Algebra Level 4

$\large \sqrt{a + \dfrac{1}{a} } + \sqrt{b+ \dfrac{1}{b}} + \sqrt{c + \dfrac{1}{c}} \geq S (\sqrt{a} + \sqrt{b} + \sqrt{c})$

Let $a,b$ and $c$ be positive numbers satisfying $ab+bc+ac=1$ . Find the largest value of $S$ for the inequality above

The answer is 2.

2 solutions

Dragan Marković
Apr 30, 2016

$a+ \frac{1}{a} =a+ \frac{ab+bc+ca}{a}=a+ b+c + \frac{bc}{a}$ Now with AM-GM: $\frac{bc}{a} + a >= 2 \sqrt{bc}$ Adding that to the first equation we get: that $a + \frac{1}{a}>= (\sqrt{b} + \sqrt{c})^2$ Similarily we get for the other ones similar result. Now by adding their square roots we get $2(\sqrt{a} + \sqrt{b} + \sqrt{c})$ Hence wanted $S=2$

By the way, $\geq$ is the better than >=.

A Former Brilliant Member - 5 years, 1 month ago

Don't have access to it on the phone so i just used >=

Dragan Marković - 5 years, 1 month ago

Use \geq within $\LaTeX$ it appears as $\geq$ .

A Former Brilliant Member - 5 years, 1 month ago

@A Former Brilliant Member – i will thanks!

Dragan Marković - 5 years, 1 month ago

Aakash Khandelwal
Apr 30, 2016

By symmetry $x=y=z=1/\sqrt{3}$

But what if they weren't equal? How would you generalise the inequality?

Dragan Marković - 5 years, 1 month ago

Another inequality

The answer is 2.

2 solutions

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