$a, b, c>0$ and $p,q,r \in [0, \displaystyle\frac{1}{2} ]$ and $a+b+c=p+q+r=1$. Prove that $8abc \leq pa+qb+rc$.
My trial was to denote $pa+qb+rc-8abc=f(p)$ and use the properties of the linear function such as the minimum point on an interval. But I could't work it out. Please help me! Thanks in advance!
On
One approach is to show that $f = pa+qb+rc-8abc \geq 0$. The value of $f$ is minimized when the value of $a$ approaches $1$ and $p = 0$. Thus, $q=r=1/2$. So, assume that $b+c = \frac{1}{n},$ where $n$ is large. Then, $$pa+qb+rc \geq 0+qb+rc = \frac{1}{2n},$$ and $$8abc \leq \frac{2}{n^2}.$$ Now, $$pa+qb+rc-8abc \geq \frac{1}{2n}-\frac{2}{n^2} \rightarrow 0^+$$ as $n \rightarrow \infty.$ So, the positivity of $f$ verifies the desired inequality.
The homogenization gives: $$(a+b+c)^2(pa+qb+rc)\geq8abc(p+q+r),$$ which is a linear inequality of $p$, of $q$ and of $r$,
which says that it's enough to prove the last inequality for $\{p,q,r\}=\{0,\frac{1}{2}\}$
Easy to see that if one number between $p$, $q$ and $r$ is equal to $0$, so two others are equal to $\frac{1}{2}$
and the case $pqr\neq0$ is impossible.
Id est, it's enough to prove our inequality in the following case:
$p=0$, $q=r$.
We need to prove that $$(a+b+c)^2(b+c)\geq16abc,$$ which is true by AM-GM: $$(a+b+c)^2(b+c)\geq\left(2\sqrt{a(b+c)}\right)^2(b+c)=4a(b+c)^2\geq4a\left(2\sqrt{bc}\right)^2=16abc.$$