<s>[INST] Inequality of area of two triangles Let $ABC$ be a triangle with sides $a,b,c$ and $A_1B_1C_1$ be another triangle with sides $a+\frac{b}2$, $b+\frac{c}2$, $c+\frac{a}2$. Prove that:
$$\frac94[ABC]\le[A_1B_1C_1]$$
I tried using Heron's formula, but it is not getting me to the answer. Can anyone help? :)
 [/INST] By Heron's formula,
$[ABC]=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$[A_1B_1C_1]=\frac{1}{4}\sqrt{\frac{3(a+b+c)}{2}(b+c-a+\frac{a+c-b}{2})(a+c-b+\frac{a+b-c}{2})(a+b-c+\frac{b+c-a}{2})}$
Set $b+c-a=x$, $a+c-b=y$, $a+b-c=z$,
Enough to show:
$\frac{27}{8}xyz\leq(x+\frac{y}{2})(y+\frac{z}{2})(z+\frac{x}{2}) $,
i.e
$27xyz\leq(x+x+y)(y+y+z)(z+z+x) $,
Use AM-GM for 3 terms of RHS.
For example, $x+x+y\geq 3\sqrt[3]{x^2y}$.
QED.
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