An Inequality in Triangle IV


An   Inequality in Triangle IV


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

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


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.


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.


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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