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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 :)

Michael Rozenberg
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Iuli
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    @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? – user 1591719 Aug 19 '12 at 13:24
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    It may be worth noting that the LHS can be written as $abcd\left(\frac{b}{a}+\frac{a}{b}+\frac{d}{c}+\frac{c}{d}\right)$ – preferred_anon Aug 19 '12 at 15:49

2 Answers2

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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 $S\le 4$.

Y.Z
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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!

Michael Rozenberg
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  • Wasnt this cool inequality proposed by you? Found it in samin riasat's handout. – Overkill2010 Oct 09 '25 at 18:53
  • @Overkill2010 Yes, it's my old problem. There is a similar(but another) very nice problem: Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a^2+b^2+c^2+d^2=4$. Prove that: $a^2bc+b^2cd+c^2da+d^2ab\leq2+2abcd$. – Michael Rozenberg Oct 09 '25 at 18:55
  • The equality occurs also for $(a,b,c,d)=(\sqrt2,1,1,0).$ By the way, I proposed more than $1000$ problems... – Michael Rozenberg Oct 09 '25 at 19:01
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    thank you for the inequality and for all the others you proposed for they are very helpful.[1000 is a lot] – Overkill2010 Oct 10 '25 at 17:12