British Maths Olympiad 1999 Round 2 Problem 3

My solution to an Olympiad problem.

By Vamshi Jandhyala in mathematics

September 27, 2019

Non-negative real numbers $p, q$ and $r$ satisfy $p+ q + r = 1$. Prove that

$$ \begin{align} 7(pq + qr + rp) \leq 2 + 9pqr. \end{align} $$

Solution

$$ \begin{align*} 2 + 9pqr &&=&& 2((p+q+r)^{3})+ 9pqr = 2(p^{3}+q^{3}+r^{3})+ 6 (p^{2}q+p^{2}r+q^{2}p+q^{2}r+r^{2}p+r^{2}q)+ 21pqr\
7(pq + qr + rp) &&=&& 7(pq+qr+rp)(p+q+r) = 7(p^{2}q+pq^{2}+pqr+pqr+q^{2}r+qr^{2}+rp^{2}+pqr+r^{2}p) \
&&=&& p^{2}q+p^{2}r+q^{2}p+q^{2}r+r^{2}p+r^{2}q + 6 (p^{2}q+p^{2}r+q^{2}p+q^{2}r+r^{2}p+r^{2}q)+ 21pqr \end{align*} $$

From Muirhead’s inequality we have

$$ \begin{align} 2(p^{3}+q^{3}+r^{3}) \geq p^{2}q+p^{2}r+q^{2}p+q^{2}r+r^{2}p+r^{2}q \end{align} $$

because $(3,0,0)$ majorizes $(2,1,0)$.

Therefore $7(pq + qr + rp) \leq 2 + 9pqr$.