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Question Number 157598 by naka3546 last updated on 25/Oct/21
Given a,b,c nonnegative numbers which satisfy a+b+c = 3. Prove that (1/(2ab^2 + 1)) + (1/(2bc^2 +1)) + (1/(2ac^2 + 1)) ≥ 1 .
$${Given}\:\:{a},{b},{c}\:\:{nonnegative}\:\:{numbers}\:\:{which}\:\:{satisfy}\:\:\:{a}+{b}+{c}\:=\:\mathrm{3}. \\ $$$${Prove}\:\:{that}\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}{ab}^{\mathrm{2}} \:+\:\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{bc}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{ac}^{\mathrm{2}} \:+\:\mathrm{1}}\:\geqslant\:\mathrm{1}\:. \\ $$
Answered by ghimisi last updated on 25/Oct/21
((abc))^(1/3) ≤((a+b+c)/3)⇒abc≤1 Σ(1/(2ab^2 +1))=Σ((((1/b))^2 )/(2a+(1/b^2 )))≥((((1/a)+(1/b)+(1/c))^2 )/(2(a+b+c)+(1/a^2 )+(1/b^2 )+(1/c^2 ))) ((((1/a)+(1/b)+(1/c))^2 )/(6+(1/a^2 )+(1/b^2 )+(1/c^2 )))≥^((•)) 1 (•)⇔(1/a^2 )+(1/b^2 )+(1/c^2 )+2((1/(ab))+(1/(bc))+(1/(ac)))≥6+(1/a^2 )+(1/b^2 )+(1/c^2 )⇔ a+b+c≥3abc⇔abc≤1
$$\sqrt[{\mathrm{3}}]{{abc}}\leqslant\frac{{a}+{b}+{c}}{\mathrm{3}}\Rightarrow{abc}\leqslant\mathrm{1} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{2}{ab}^{\mathrm{2}} +\mathrm{1}}=\Sigma\frac{\left(\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} }{\mathrm{2}{a}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}\geqslant\frac{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}+{c}\right)+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }} \\ $$$$\frac{\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)^{\mathrm{2}} }{\mathrm{6}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }}\overset{\left(\bullet\right)} {\geqslant}\mathrm{1} \\ $$$$\left(\bullet\right)\Leftrightarrow\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }+\mathrm{2}\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ac}}\right)\geqslant\mathrm{6}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\Leftrightarrow \\ $$$${a}+{b}+{c}\geqslant\mathrm{3}{abc}\Leftrightarrow{abc}\leqslant\mathrm{1} \\ $$
Commented by naka3546 last updated on 25/Oct/21
Thank you, sir.
$${Thank}\:\:{you},\:\:{sir}. \\ $$

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