How can I prove this multilinear inequality?

$$a, b, c>0$$ 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!

• Where is the linearity ? – Yves Daoust Jun 26 at 17:18
• It can be considered a linear function, which makes it a linear inequality. – furfur Jun 26 at 17:18
• Which "it" do you mean ? I see two affine constraints and a nonlinear inequality. – Yves Daoust Jun 26 at 17:19
• The expression pa+qb+rc-8abc=a(p-8bc)+(qb+rc) can be considered a linear function in a. – furfur Jun 26 at 17:20
• Thank you for telling me! – furfur Jun 26 at 17:22

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: $$(a+b+c)^2(b+c)\geq\left(2\sqrt{a(b+c)}\right)^2(b+c)=4a(b+c)^2\geq4a\left(2\sqrt{bc}\right)^2=16abc.$$

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.