6
$\begingroup$

Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$

Prove the inequality

$$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$

Thanks :)

$\endgroup$
2
  • 2
    $\begingroup$ @Iuli: as you already know, the inequalities may be a painful thing since some of them require a lot of time to think. That's why it helps mention anything you know about it, what you've already tried. You have (+1) for the question. Could Cauchy-Schwarz help? $\endgroup$ Commented Aug 19, 2012 at 13:24
  • 1
    $\begingroup$ It may be worth noting that the LHS can be written as $abcd\left(\frac{b}{a}+\frac{a}{b}+\frac{d}{c}+\frac{c}{d}\right)$ $\endgroup$ Commented Aug 19, 2012 at 15:49

2 Answers 2

12
$\begingroup$

Let $S=a^2bc+b^2cd+c^2da+d^2ab$. We can easily find that:

$$S-(ac+bd)(ab+cd)=-bd(a-c)(b-d);$$ $$S-(bc+ad)(bd+ac)=ac(a-c)(b-d)$$ which implies $$S\le \max\{(ac+bd)(ab+cd),(bc+ad)(bd+ac)\}.$$

By AG mean inequality:

\begin{align*} (ac+bd)(ab+cd)&\le \left(\frac{(ac+bd)+(ab+cd)}{2}\right)^2\\ {}&=\frac{(a+d)^2(b+c)^2}{4}\\ {}&\le \frac{1}{4}\left[\left(\frac{a+d+b+c}{2}\right)^2\right]^2\\ {}&=4 \end{align*} Similarly, we have $$(bc+ad)(bd+ac)\le 4$$.

Thus we have $S\le 4$.

$\endgroup$
0
1
$\begingroup$

Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.

Hence, since $(x,y,z,t)$ and $(xyz,xyt,xzt,yzt)$ are the same ordered,

by Rearrangement and AM-GM we obtain: $$a^2bc+b^2cd+c^2da+d^2ab=a\cdot abc+b\cdot bcd+c\cdot cda+d\cdot dab\leq$$ $$\leq x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt=xy(xz+yt)+zt(xz+yt)=$$ $$=(xy+zt)(xz+yt)\leq\left(\frac{xy+xz+zt+yt}{2}\right)^2=$$ $$=\left(\frac{(x+t)(y+z)}{2}\right)^2\leq\left(\frac{\left(\frac{x+y+z+t}{2}\right)^2}{2}\right)^2=4.$$ Done!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .