An inequality problem by Gurido Cuong

We call the expression L, we see that $c+3a-d\leq c+3a \Rightarrow \frac{1}{\sqrt[3]{c+3a-d}}\geq\frac{1}{\sqrt[3]{c+3a}}$ $12abcd\geq 0$ Now we choose $A=\frac{1}{\sqrt[3]{a+2b}}+\frac{1}{\sqrt[3]{b+2c}}+\frac{1}{\sqrt[3]{c+2a}}$ $B= (a+2b)+(b+2c)+(c+2a)=3$ Applying Holder's Inequality $A^3B\geq 3^4$ $\Leftrightarrow A\geq 3$ With $-24abc$ we easily applying AM-GM to get $-24abc\geq -24\big(\frac{a+b+c}{3}\big)^3=-24.4^{-3}$

Finally $L\geq 3-24.4^{-3}=2.625$ The equality holds when $a=b=c=\frac{1}{4}$ and $d=0$

áp dụng cauchy 3 số cho 3 hạng tử đầu đúng k bạn?