Question

State true or False.

If a, b, c are sides of a triangle, then $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

• A. True
• B. False

Answer 1

The correct option is A True

$\frac{a+a+(b+c-a)}{3}\ge {\left[a^2\right(b+c-a\left)\right]}^{1/3}{(a+b+c)}^3\ge 27\left[a^2\right(b+c-a\left)\right]\left(1\right)\frac{b+b+(c+a-b)}{3}\ge {\left[b^2\right(c+a-b\left)\right]}^{1/3}{(a+b+c)}^3\ge 27\left[b^2\right(c+a-b\left)\right]\left(2\right)\frac{c+c+a+b-c}{3}\ge {\left[c^2\right(b+c-a\left)\right]}^{1/3}{(a+b+c)}^3\ge 27\left[c^2\right(a+b-c\left)\right]\left(3\right)$

Adding $\left(1\right),\left(2\right)$ and $\left(3\right)$

$3{(a+b+c)}^3\ge 27\left[a^2\right(b+c-a)+b^2(c+a-b)+c^2(a+b-c\left)\right]{(a+b+c)}^3\ge 9\left[a^2\right(b+c-a)+b^2(c+a-b)+c^2(a+b-c\left)\right]\left(4\right)$

Also, $\frac{(a+b+c)}{3}\ge {\left(abc\right)}^{1/3}{(a+b+c)}^3\ge 27\left(abc\right)\left(5\right)$

Dividing $\left(4\right)$ by $\left(5\right)$

$1\ge \frac{a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)}{3\left(abc\right)}3abc\ge a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)$