An Inequality in Triangle IV

Problem

An   Inequality in Triangle IV

Lemma

$\sqrt{p(p-b)}+\sqrt{p(p-c)}\le \sqrt{2(p^2-m_a^2)}.$

Set $a=y+z,\;$ $b=x+z,\;$ $c=x+y.\;$ Then $p=x+y+z,\;$ $y=p-b,\;$ and $z=p-c,\;$ and, since $\displaystyle m_a^2=\frac{b^2+c^2}{2}-\frac{a^2}{4},\;$ $\displaystyle m_a^2=\frac{4x(x+y+z)+(y-z)^2}{4}.$

Thus the lemma inequality is equivalent to

$\displaystyle\sqrt{x+y+z}(\sqrt{y}+\sqrt{z})\le\sqrt{2(x+y+z)(y+z)-\frac{(y-z)^2}{2}},$

which reduces to

$\displaystyle\frac{(\sqrt{y}-\sqrt{z})^2(\sqrt{y}+\sqrt{z})^2}{2}\le (x+y+z)(\sqrt{y}-\sqrt{z})^2$

and, subsequently, to

$\displaystyle\frac{(\sqrt{y}+\sqrt{z})^2}{2}\le (x+y+z).$

But this is immediate.

Proof

Using the lemma,

$\displaystyle\begin{align}\sqrt{p(p-b)}+\sqrt{p(p-c)}\le \sqrt{2(p^2-m_a^2)},\\ \sqrt{p(p-c)}+\sqrt{p(p-a)}\le \sqrt{2(p^2-m_b^2)},\\ \sqrt{p(p-a)}+\sqrt{p(p-b)}\le \sqrt{2(p^2-m_c^2)}. \end{align}$

Adding the three gives the desired result.

Acknowledgment

The problem which is due to the Romanian school teacher Constantin Tanasescu has been posted at the CutTheKnotMath facebook page by Leo Giugiuc, along with a proof by Leo Giugiuc and Dan Sitaru.

 

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