den_thewhitelion wrote: a,b,c>0 and $abc\ge 1$.Prove that: $\dfrac{1}{a^3+2b^3+6}+\dfrac{1}{b^3+2c^3+6}+\dfrac{1}{c^3+2a^3+6} \le \dfrac{1}{3}$ only need to prove it with abc=1 which is easy

Can you prove it please? P.S:the problem is really easy but you need that abc$\ge1$ only at the end of the proof

$\sum \frac{1}{a^3+2b^3+6}=\sum \frac{1}{a^3++b^3+b^3+6} \le \sum \frac{1}{3ab^2+6}=\sum \frac{1}{3(ab^2+2)}$ so it suffices to prove $ \sum \frac{1}{ab^2+2} \le 1$ $\iff$ $ \sum \frac{2}{ab^2+2} \le 2$ $\iff$ $\sum \frac{ab^2}{ab^2+2} \ge 1$ but $2 \le 2abc$ so it suffices to prove $\sum \frac{ab^2}{ab^2+2abc} \ge 1$ $\iff$ $\sum \frac{b}{b+2c} \ge 1$ $\iff$ $\sum \frac{b^2}{b^2+2bc} \ge 1$ which is just C-S (cause $\sum \frac{b^2}{b^2+2bc} \ge \frac{(a+b+c)^2}{(a+b+c)^2}= 1$ )

Let $a,b,c>0$. Prove that: $$\dfrac{1}{a^3+2b^3+3abc}+\dfrac{1}{b^3+2c^3+3abc}+\dfrac{1}{c^3+2a^3+3abc} \le \dfrac{1}{2abc}$$

Lemma: $\frac{abc}{a^3+2b^3+6abc} \leq \frac{a+2c}{9(a+b+c)}.$ Proof: It suffices to show $a^4+2ab^3+6a^2bc+2a^3c+4b^3c+12abc^2 \geq 9a^2bc+9ab^2c+9abc^2$ $\iff a^4+2ab^3+2a^3c+4b^3c+3abc^2 \geq 3a^2bc+9ab^2c$ but we have the following by AM-GM: $$a^3c + a^3c+b^3c \geq 3a^2bc \qquad \text{and} \qquad a^4+ab^3+ab^3+b^3c+b^3c+b^3c+abc^2+abc^2+abc^2 \geq 9ab^2c.$$ The inequality above follows by adding these two inequalities, so the lemma is proven. As the post above says, it suffices to prove the inequality of the problem for $abc=1$. Using the lemma, we are done.

Achillys wrote: Lemma: $\frac{abc}{a^3+2b^3+6abc} \leq \frac{a+2c}{9(a+b+c)}.$ Proof: It suffices to show $a^4+2ab^3+6a^2bc+2a^3c+4b^3c+12abc^2 \geq 9a^2bc+9ab^2c+9abc^2$ $\iff a^4+2ab^3+2a^3c+4b^3c+3abc^2 \geq 3a^2bc+9ab^2c$ but we have the following by AM-GM: $$a^3c + a^3c+b^3c \geq 3a^2bc \qquad \text{and} \qquad a^4+ab^3+ab^3+b^3c+b^3c+b^3c+abc^2+abc^2+abc^2 \geq 9ab^2c.$$ The inequality above follows by adding these two inequalities, so the lemma is proven. As the post above says, it suffices to prove the inequality of the problem for $abc=1$. Using the lemma, we are done. Excuse me. How does the lemma come to you @@?(sorry for the poor English and poor math.)

andysteph wrote: Excuse me. How does the lemma come to you @@?(sorry for the poor English and poor math.) It's isolated fudging, but a single variable won't work on the numerator of RHS, so it also takes a bit of trial and error.

den_thewhitelion wrote: a,b,c>0 and $abc\ge 1$.Prove that: $\dfrac{1}{a^3+2b^3+6}+\dfrac{1}{b^3+2c^3+6}+\dfrac{1}{c^3+2a^3+6} \le \dfrac{1}{3}$

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Achillys wrote: andysteph wrote: Excuse me. How does the lemma come to you @@?(sorry for the poor English and poor math.) It's isolated fudging, but a single variable won't work on the numerator of RHS, so it also takes a bit of trial and error. Did you just try random values for the numerator after finding the exponent?

den_thewhitelion wrote: a,b,c>0 and $abc\ge 1$.Prove that: $$\dfrac{1}{a^3+2b^3+6}+\dfrac{1}{b^3+2c^3+6}+\dfrac{1}{c^3+2a^3+6} \le \dfrac{1}{3}$$ here Let $a,b,c$ be positive real numbers with $abc=1$. Prove the inequality $$\frac{1}{a^3+2b^3+3}+\frac{1}{b^3+2c^3+3}+\frac{1}{c^3+2a^3+3}\leq\frac{1}{2} $$

SidVicious wrote: $\sum \frac{1}{a^3+2b^3+6}=\sum \frac{1}{a^3++b^3+b^3+6} \le \sum \frac{1}{3ab^2+6}=\sum \frac{1}{3(ab^2+2)}$ so it suffices to prove $ \sum \frac{1}{ab^2+2} \le 1$ $\iff$ $ \sum \frac{2}{ab^2+2} \le 2$ $\iff$ $\sum \frac{ab^2}{ab^2+2} \ge 1$ but $2 \le 2abc$ so it suffices to prove $\sum \frac{ab^2}{ab^2+2abc} \ge 1$ $\iff$ $\sum \frac{b}{b+2c} \ge 1$ $\iff$ $\sum \frac{b^2}{b^2+2bc} \ge 1$ which is just C-S (cause $\sum \frac{b^2}{b^2+2bc} \ge \frac{(a+b+c)^2}{(a+b+c)^2}= 1$ ) simple ending: after $ \sum \frac{1}{ab^2+2} \le 1$ it would suffice to prove that: 3 ≤ a^2b+c^2a+b^2c which follows directly by AM-GM

\[\sum{\frac{1}{(b^3+c^3+1)+(c^3+1+1)+3}}\leq \frac{1}{3}\sum{\frac{1}{bc+c+1}}\]\[\sum{\frac{1}{bc+c+1}}=\frac{1}{bc+c+1}+\frac{bc}{abc^2+abc+bc}+\frac{c}{abc+bc+c}\leq \frac{1}{bc+c+1}+\frac{bc}{bc+c+1}+\frac{c}{bc+c+1}=1\]As desired.$\blacksquare$