Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$

Question

Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that,

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

Well from their sum we do get that $\frac{1}{4} \geq \sqrt[4]{abcd}$

$$\Rightarrow \frac{1}{4^8} \geq a^2b^2c^2d^2$$

and applying AM-GM on LHS of the given inequality we get ,

$$\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot \sqrt[4]{\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)}}$$

and $(a+b)(b+c)(c+d)(d+a) \geq 16 \cdot abcd$ or

$$\frac{a^3b^3c^3d^3}{(a+b)(b+c)(c+d)(d+a)} \leq 16 \cdot a^2b^2c^2d^2 \leq 4^2 \cdot \frac{1}{4^8}= \frac{1}{4^6}$$

$$\Rightarrow \frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq 4 \cdot\frac{1}{8}=\frac{1}{2} > \frac{1}{8} \blacksquare.$$

Is this proof correct ? Did i miss any details? My doubt really stems from the fact that i didnt get $\frac{1}{8}$ directly but $\frac{1}{2}$, which makes my resultant inequality strict instead of being $\geq$ and it makes me wonder whether my proof is right. Thanks.

EDIT: Well guys i haven't read Titu's Lemma or Holder's inequality just yet though both of them do seem very powerful. I guess i'll just come to this question later when m done with those topics. Thanks for your help. Also I was just wondering whether it is possible to do it purely using AM-GM or maybe WAM-WGM ? Thanks again.

In your proof, what you have is $A \geq B$ and $ B \leq C$ , from which you concluded $ A \geq C$. Do you spot an error? — Calvin Lin, Jun 07 '20 at 11:32
As to how to solve the problem, hint: Apply Titu's lemma to $ \sum \frac{a^4}{ab+ac}$. The rest follows almost immediately. — Calvin Lin, Jun 07 '20 at 11:49
Does this answer your question? a mathematical problem on inequalities — Itoz Darbien, Jul 02 '23 at 06:50

score 3 · Answer 1 · answered Jun 07 '20 at 12:14

Credits to Calvin Lin for the hint to apply Titu's Lemma, but I don't find the remaining part of the proof obvious.

Applying Titu's lemma yields: \begin{align*} \frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} &= \frac{a^4}{ab+ac}+\frac{b^4}{bc+bd}+\frac{c^4}{cd+ca}+\frac{d^4}{da+db} \\ &\geq \frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db} \end{align*} Applying Cauchy-Schwarz Inequality to the denominator yields: \begin{align*} ab + ac + bc + bd + cd + ca + da + db &\leq \sqrt{\left(2a^2 + 2b^2 + 2c^2 + 2d^2 \right)^2} \\ &= 2(a^2 + b^2 + c^2 + d^2) \end{align*} Therefore: \begin{align*} \frac{(a^2 + b^2 + c^2 + d^2)^2}{ab + ac + bc + bd + cd + ca + da + db} \geq \frac{1}{2}(a^2 + b^2 + c^2 +d^2) \end{align*} Applying Cauchy-Schwarz inequality again with $(a^2,b^2,c^2,d^2) \cdot (1,1,1,1)$ yields:: \begin{align*} 4(a^2 + b^2 + c^2 + d^2) \geq (a + b + c + d)^2 = 1 \end{align*} Therefore: \begin{align*} \frac{1}{2}(a^2 + b^2 + c^2 +d^2) \geq \frac{1}{2}\left(\frac{1}{4}\right) = \frac{1}{8} \end{align*}

Hey , how to prove the above using pure AM-GM-HM ? – Ash_Blanc Jul 02 '23 at 02:07 — Ash_Blanc, Jul 02 '23 at 02:07

Michael Rozenberg · Answer 2 · 2020-06-07T13:07:54.693

2

No. Your proof is not correct.

After your first step we need to prove a wrong for $a\rightarrow0^+$ inequality.

I think it's better to use Holder: $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c)}=\frac{1}{8}.$$

edited Jun 07 '20 at 13:07

answered Jun 07 '20 at 13:00

Michael Rozenberg

194,933

sir is there any way to prove the above using purely A.M.-G.M.-H.M. ? – Ash_Blanc Jul 02 '23 at 02:09

Let $a, b, c, d \in R^+$ such that $a + b + c + d = 1$. Prove that $\frac{a^3}{b+c}+\frac{b^3}{c+d}+\frac{c^3}{d+a}+\frac{d^3}{a+b} \geq \frac{1}{8}$

2 Answers2