$$\left(\sum\frac{a}{\sqrt[3]{a+2b+c}}\right)^3\left(\sum a^2+2ab+ac\right)\geq (a+b+c+d)^4$$$$\iff \left(\sum\frac{a}{\sqrt[3]{a+2b+c}}\right)^3\cdot 8^2\geq 8^4$$$$\iff \sum\frac{a}{\sqrt[3]{8+b-d}}\geq 4$$by Holder.

gobathegreat wrote: Let $a$, $b$, $c$ and $d$ be real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$ Use Jensen's inequality for the convex function $f(x)=\frac{1}{\sqrt[3]{8+x}}$: $$\frac a 8 f(b-d)+\frac b 8 f(c-a)+\frac c 8 f(d-b)+\frac d 8 f(a-c)\geq f\left(\frac a 8 (b-d)+\frac b 8 (c-a)+\frac c 8 (d-b)+\frac d 8 (a-c)\right)=f(0)=\frac 1 2.$$

gobathegreat wrote: Let $a$, $b$, $c$ and $d$ be real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$ It's wrong. Try $a=-1$ and $b-d\rightarrow-8^+$.

arqady wrote: gobathegreat wrote: Let $a$, $b$, $c$ and $d$ be real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$ It's wrong. Try $a=-1$ and $b-d\rightarrow-8^+$. I think we should take $a$, $b$, $c$, $d>0$.

PavelMath wrote: I think we should take $a$, $b$, $c$, $d>0$. Why? It's not given.

gobathegreat wrote: Let $a$, $b$, $c$ and $d$ be real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$ Let $a$, $b$, $c$ and $d$ be positive real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$Proof of Zhangyunhua:

Attachments:

Let $a$, $b$, $c$ and $d$ benonnegative real numbers such that $a+b+c+d=8$. Prove the inequalityFind the maximum value of $\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{d+3}}+\sqrt{\frac{d}{a+3}}$

sqing wrote: Let $a$, $b$, $c$ and $d$ benonnegative real numbers such that $a+b+c+d=8$. Prove the inequalityFind the maximum value of $\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{d+3}}+\sqrt{\frac{d}{a+3}}$ this reminds me of IMOSL 2018 A7

In the Israel's last test was the following. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$

arqady wrote: In the Israel's last test was the following. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$ Beautiful. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=100.$ Prove that: $$\sqrt[3]{\frac{a}{b+7}} + \sqrt[3]{\frac{b}{c+7}} + \sqrt[3]{\frac{c}{d+7}} + \sqrt[3]{\frac{d}{a+7}} \leq\frac{8}{\sqrt[3]7}.$$

sqing wrote: gobathegreat wrote: Let $a$, $b$, $c$ and $d$ be real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$ Let $a$, $b$, $c$ and $d$ be positive real numbers such that $a+b+c+d=8$. Prove the inequality: $$\frac{a}{\sqrt[3]{8+b-d}}+\frac{b}{\sqrt[3]{8+c-a}}+\frac{c}{\sqrt[3]{8+d-b}}+\frac{d}{\sqrt[3]{8+a-c}} \geq 4$$Proof of Zhangyunhua: to sqing in your solutions, how to get statement 2->3? in advanced thank you

arqady wrote: In the Israel's last test was the following. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$ @arqady whats the solution?

shalomrav wrote: arqady wrote: In the Israel's last test was the following. Let $a$, $b$, $c$ and $d$ be non-negative numbers such that $a+b+c+d=18.$ Prove that: $$\sqrt{\frac{a}{b+6}}+\sqrt{\frac{b}{c+6}}+\sqrt{\frac{c}{d+6}}+\sqrt{\frac{d}{a+6}}\leq5\sqrt{\frac{2}{7}}$$ @arqady whats the solution? Here's an ugly way, hopefully someone can find a better solution. By rearrangement inequality if we sort $a,b,c,d$ into $u,v,w,x$ in ascending order, we obtain $$LHS \le \sqrt{\frac{u}{w+6}} + \sqrt{\frac{w}{u+6}} + \sqrt{\frac{v}{x+6}} + \sqrt{\frac{x}{v+6}}$$and we have $$\sqrt{\frac{w}{u+6}} + \sqrt{\frac{u}{w+6}} = \sqrt{\frac{w(u+6)}{(u+6)^2}} + \sqrt{\frac{u(w+6)}{(w+6)^2}} \le \sqrt{(w(u+6) + u(w+6))(1/(w+6)^2 + 1/(u+6)^2)} = \frac{\sqrt{(uw + 3k)((k+12)^2 + k^2 - 4uw)}}{uw + 6k + 36}$$where $k = u+w$. Taking derivatives show that the maximum is achieved when $uw = \frac{216}{18+k}$, and this maximum is $\frac{k+6}{\sqrt{6(k+12)}}$. Another derivative now gives that for any $k,l > 0$ with $k+l = 18$ we have $$\frac{k+6}{\sqrt{6(k+12)}} + \frac{l+6}{\sqrt{6(l+12)}} \le \frac{5\sqrt{2}}{7}$$as desired.

HolyMath wrote: to sqing in your solutions, how to get statement 2->3? in advanced thank you $\frac{(a+b+c+d)^2}{a\sqrt[3]{8+b-d}+b\sqrt[3]{8+c-a}+a\sqrt[3]{8+d-b}+a\sqrt[3]{8+a-c}}$ $=\frac{4(a+b+c+d)^2}{a \cdot 2 \cdot 2\sqrt[3]{8+b-d}+b \cdot 2 \cdot 2 \sqrt[3]{8+c-a}+c \cdot 2 \cdot 2 \sqrt[3]{8+d-b}+d \cdot 2 \cdot 2 \sqrt[3]{8+a-c}} $ $\geq \frac{4(a+b+c+d)^2}{a \frac{2^3+2^3+8+b-d}{3}+b\frac{2^3+2^3+8+c-a}{3}+c \frac{2^3+2^3+8+d-b}{3}+d \frac{2^3+2^3+8+a-c}{3}}$ (By $AM-GM$) $=\frac{4(a+b+c+d)^2}{a \frac{24+b-d}{3}+b\frac{24+c-a}{3}+c \frac{24+d-b}{3}+d \frac{24+a-c}{3}}$.