# 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.