Question

For a>0, b>0, c>0, which of the following hold good?

• A. $2(a^3+b^3+c^3)\ge bc(b+c)+ca(c+a)+ab(a+b)$
• B. $\frac{a^3+b^3+c^3}{3}>\frac{a+b+c}{3}.\frac{a^2+b^2+c^2}{3}$
• C. $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{1}{2}(a+b+c)$
• D. $\frac{2}{b+c}+\frac{2}{c+a}+\frac{2}{a+b}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $2(a^3+b^3+c^3)\ge bc(b+c)+ca(c+a)+ab(a+b)$
B $\frac{a^3+b^3+c^3}{3}>\frac{a+b+c}{3}.\frac{a^2+b^2+c^2}{3}$
C $\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{1}{2}(a+b+c)$
D $\frac{2}{b+c}+\frac{2}{c+a}+\frac{2}{a+b}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

(a), (b), (c), (d)
(a) $a^2+b^2\ge 2abor{a}^2-ab+b^2\ge ab$
$\therefore (a+b)(a^2-ab+b^2)\ge ab(a+b)$
or $a^3+b^3\ge ab(a+b)$
Similarly $b^3+c^3\ge bc(b+c)$
and $c^3+a^3\ge ca(c+a)$
whence by addition, we get
$2(a^3+b^3+c^3)>ab(a+b)+bc(b+c)+ca(c+a)$

(b) We have to prove that
$3(a^3+b^3+c^3)>(a^3+b^3+c^3)+\sum a(b^2+c^2)$
or $2(a^3+b^3+c^3)>\sum ab(a+b)$
we have proved the above in part (a)
(c,d) We know that $(b+c)>2\sqrt{bc}$
Square both sides
$\therefore 4bc<(b+c{)}^2$
or $\frac{4bc}{b+c}<(b+c)$, for (a)
and $\frac{4}{b+c}<\frac{b+c}{bc}$ or $<\frac{1}{b}+\frac{1}{c}$, for (b)
Similarly write other similar inequalities and add.