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# Inequality 40(Murray Klamkin)

Problem:

If $\displaystyle x,y,z>0$ prove that

$\displaystyle \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\geq 3\sqrt{xy+yz+zx}$.

Solution:

From the well-known lemma $\displaystyle 4(a^2+ab+b^2)\geq 3(a+b)^2$ we deduce that:

$\displaystyle 2\sqrt{a^2+ab+b^2}\geq \sqrt{3}(a+b)$.

Doing that cyclic for $\displaystyle x,y,z$ and adding up the $\displaystyle 3$ relations we get that

$\displaystyle 2\sum_{cyc}\sqrt{x^2+xy+y^2}\geq \sqrt{3}\cdot 2\sum_{cyc}x\Longrightarrow \sum\sqrt{x^2+xy+y^2}\geq \sqrt{3}\sum_{cyc}x$.

So, it is enough to prove that

$\displaystyle \sqrt{3}\sum_{cyc}x\geq 3\sqrt{xy+yz+zx}$.

Squaring both sides we come to the conclusion

$\displaystyle \left(\sum_{cyc}x\right)^{2}\geq 3\sum_{cyc}xy$, Q.E.D.