1
$\begingroup$

$a, b, c>0$ 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!

$\endgroup$
  • $\begingroup$ Where is the linearity ? $\endgroup$ – Yves Daoust Jun 26 at 17:18
  • $\begingroup$ It can be considered a linear function, which makes it a linear inequality. $\endgroup$ – furfur Jun 26 at 17:18
  • $\begingroup$ Which "it" do you mean ? I see two affine constraints and a nonlinear inequality. $\endgroup$ – Yves Daoust Jun 26 at 17:19
  • $\begingroup$ The expression pa+qb+rc-8abc=a(p-8bc)+(qb+rc) can be considered a linear function in a. $\endgroup$ – furfur Jun 26 at 17:20
  • 1
    $\begingroup$ Thank you for telling me! $\endgroup$ – furfur Jun 26 at 17:22
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.