<s>[INST] Prove that: $\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$ Problem: Let $a,b,c>0; k\ge1$. Prove that: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{6}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$$
My attempt: Using C-S inequality: $$\frac{1}{ka+b}+\frac{1}{kb+c}+\frac{1}{kc+a}\ge\frac{9}{(k+1)(a+b+c)}$$
It is desired to prove: $$\sqrt[3]{(a+b)(b+c)(c+a)}\ge\frac{2}{3}(a+b+c)$$ Which is false.
The problem has nice form but for me, it is quite hard to get nice proof. Please help me, thanks!
 [/INST] I just get a part by AM-GM. Maybe stronger tools can help to make up the rest $1<k<2$
For $k=1$: the inequality holds obviously by AM-GM
For $k\ge2$: Using C-S inequality: $$k(a+b+c)\sum{\frac{1}{ka+b}}=3+(k-1)\sum{\frac{b}{ka+b}}+k\sum{\frac{c}{ka+b}}\ge\frac{6k}{k+1}+\frac{k(a+b+c)^2}{(k+1)(ab+bc+ca)}$$Also by AM-GM inequality and well- known result:  $9(a+b)(b+c)(c+a)\ge8(ab+bc+ca)(a+b+c)$, we obtain:
$$2.\frac{3k}{k+1}+\frac{k(a+b+c)^2}{(k+1)(ab+bc+ca)}\ge3\sqrt[3]{\frac{9k^3}{(k+1)^3}.\frac{(a+b+c)^2}{ab+bc+ca}}\ge\frac{6k(a+b+c)}{(k+1)\sqrt[3]{(a+b)(b+c)(c+a)}}$$Hence, the proof is cleared!
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