Home » Uncategorized » Inequality 16(Vasile Cirtoaje)

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.

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