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Daca \( a,b,c\in (0,\infty) \) astfel incat \( ab+bc+ca=1 \) atunci :

\( 9a^2b^2c^2+abc(\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{1+c^2})\le 1 \)

Gh. Stoica ,rev. Arhimede1-6,2009

\( \sqrt{1+a^2}=\sqrt{ab+bc+ca+a^2}=\sqrt{(a+c)(b+a) \)
\( \sqrt{1+b^2}=\sqrt{ab+bc+ca+b^2}=\sqrt{(b+c)(b+a) \)
\( \sqrt{1+c^2}=\sqrt{ab+bc+ca+c^2}=\sqrt{(c+a)(b+c)} \)
Sa notam cu \( E=9a^2b^2c^2+abc(\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{1+c^2}) \)
\( E=9a^2b^2c^2+abc(\sqrt{(a+c)(b+a)}+\sqrt{(c+a)(b+c)}+\sqrt{(b+c)(a+b)} \)
\( E\le 9a^2b^2c^2+2abc(a+b+c) \) (*);
Din Inegalitatea mediilor: \( \sqrt[3]{(abc)^2}\le \frac{ab+bc+ca}{3} => 9a^2b^2c^2\le \frac{(ab+bc+ca)^3}{3} =\frac{1}{3} \) (1)
In continuare folosim inegalitatea :\( (ab+bc+ca)^2\ge 3abc(a+b+c) => abc(a+b+c)\le \frac{1}{3} \) (2)
Din relatiile (*),(1),(2) \( => E\le \frac{1}{3}+2* \frac{1}{3}=1 \),q.e.d.