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# Inequality 16(Vasile Cirtoaje)

Problem:

If $\displaystyle a,b,c$ are positive real numbers, prove that

$\displaystyle \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 1+\sqrt{1+\sqrt{(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}}$.

Solution:

We know that

$\displaystyle (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\sqrt{\left(\sum_{cyc}a^2+2\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}+2\sum_{cyc}\frac{1}{ab}\right)}$.

So applying Cauchy-Schwartz inequality we have that:

$\displaystyle \sqrt{\left(\sum_{cyc}a+2\sum_{cyc}ab\right)\left(\sum_{cyc}\frac{1}{a^2}+2\sum_{cyc}\frac{1}{ab}\right)}\geq \sqrt{\sum_{cyc}a^2\sum_{cyc}\frac{1}{a^2}}+2\sqrt{\sum_{cyc}ab\sum_{cyc}\frac{1}{ab}}$,

or the right hand side can be rewritten as

$\displaystyle \sqrt{\sum_{cyc}a^2\sum_{cyc}\frac{1}{a^2}}+2\sqrt{\sum_{cyc}a\sum_{cyc}\frac{1}{a}}$

by both dividing and multiplying with $\displaystyle abc$.

Bringing now the second term of the right hand side at the left hand side and adding up $\displaystyle 1$ in both sides we get that:

$\displaystyle \left(\sqrt{\sum_{cyc}a\sum_{cyc}\frac{1}{a}}-1\right)^{2}\geq 1+\sqrt{\sum_{cyc}a^2\sum_{cyc}\frac{1}{a^2}}$,

or

$\displaystyle \sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 1+\sqrt{1+\sqrt{(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}}$, Q.E.D.