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A while ago, the following question was posted by the OP, who requested for help in proving the following inequality:

Let $a,b,c,d$ be positive reals such that $a+b+c+d=1$. Prove that:

$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{a+d}+\frac{d^3}{a+b} \ge \frac{1}{8}.$$

This was the original problem as stated by the OP.

I have added the extra condition that $a,b,c,d$ must all be positive reals, since the given inequality by itself would not hold otherwise, as another user has pointed out in the comments. Since the original post has been closed (in all fairness to the OP, the condition of positive reals could be implied, especially if this was a competition math problem), I decided to open a new post in order to provide a possible solution for the OP, as well as inquire if there are other possible means to prove this inequality. My solution involves AM-GM, so it would be nice to see other methods that work as well.

Michael Rozenberg
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Hello
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3 Answers3

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The cubic term in the numerators of each term in the L.H.S. is highly suggestive that AM-GM could be applied in some way. Hence, equivalently, we shall prove the following:

$$\sum_{cyc}\left(\frac{a^3}{b+c}+\frac{1}{32}\right)\ge \frac{1}{4}.$$

By AM-GM, we have that:

$$\frac{a^3}{b+c}+\frac{b+c}{16}+\frac{1}{32} \ge 3 \sqrt[3]{\frac{a^3}{512}}=\frac{3a}{8}$$ $$\Rightarrow \frac{a^3}{b+c} + \frac{1}{32} \ge \frac{3a}{8}-\frac{b+c}{16}$$ $$\Rightarrow \sum_{cyc}\left(\frac{a^3}{b+c}+\frac{1}{32}\right) \ge \frac{3(a+b+c+d)}{8}-\frac{2(a+b+c+d)}{16}$$ By the given condition $a+b+c+d=1$, we thus have that: $$\sum_{cyc}\left(\frac{a^3}{b+c}+\frac{1}{32}\right) \geq \frac{3}{8}-\frac{1}{8} = \frac{1}{4}.$$

Hence, this concludes our proof.

Hello
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  • Hi, this is a really nice proof. But can you please tell me how you figured out what to apply AM-GM to? like i get that because of the $a^3$ term, we would have to use AMGM on 3 terms, on of which would be like $\frac{b+c}{k}$ for some $k$. But how did you figure out k=16 and the other term is 32? Thanks! – Aditya_math Aug 28 '21 at 15:34
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Holder helps: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$

Michael Rozenberg
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Also, by C-S and AM-GM we obtain: $$\sum_{cyc}\frac{a^3}{b+c}=\sum_{cyc}\frac{a^4}{ab+ac}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}(ab+ac)}=$$ $$=\frac{(a^2+b^2+c^2+d^2)^2}{(a+c)(b+d)+2(ac+bd)}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\left(\frac{a+c+b+d}{2}\right)^2+a^2+c^2+b^2+d^2}=$$ $$=\frac{(a^2+b^2+c^2+d^2)^2}{\frac{1}{4}+a^2+c^2+b^2+d^2}.$$ Id est, it's enough to prove that: $$\frac{(a^2+b^2+c^2+d^2)^2}{\frac{1}{4}+a^2+c^2+b^2+d^2}\geq\frac{1}{8}$$ or $$32(a^2+b^2+c^2+d^2)^2-4(a^2+b^2+c^2+d^2)-1\geq0$$ or $$(4(a^2+b^2+c^2+d^2)-1)(8(a^2+b^2+c^2+d^2)+1)\geq0$$ or $$4(a^2+b^2+c^2+d^2)\geq1,$$ which is true by C-S: $$4(a^2+b^2+c^2+d^2)=\sum_{cyc}1^2\sum_{cyc}a^2\geq(a+b+c+d)^2=1.$$

Michael Rozenberg
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