Fie a,b,c>0

[Marius Mainea]

Fie a,b,c>0 cu a+b+c=1. Demonstrati:

\( a\sqrt{1-bc}+b\sqrt{1-ca}+c\sqrt{1-ab}\geq \frac{2\sqrt2}{3} \)

Mathlinks 2008

[Mateescu Constantin]

Aplicam inegalitatea lui Cebasev secventelor la fel ordonate \( \left\[\ \begin{array}{cc}
a & \ge & b & \ge & c \\\\\\\\\\
\sqrt{1-bc} & \ge & \sqrt{1-ca} & \ge & \sqrt{1-ab}\ \end{array}\right\]\ : \)

\( \sum a\sqrt{1-bc}\ge\frac 13(a+b+c)\(\sum\sqrt{1-bc}\)=\sum\frac{\sqrt{1-bc}}{3} \). Prin urmare este suficient sa aratam ca \( \sum\sqrt{1-bc}\ \ge\ 2\sqrt 2 \).

Aplicand \( AM-GM \) obtinem \( abc\le\left(\frac{a+b+c}3\right)^3=\frac 1{27} \) si \( (1-a)(1-b)(1-c)\le\left(\frac{3-a-b-c}3\right)^3 = \frac 8{27} \)

Pe de alta parte avem

\( (1-ab)(1-bc)(1-ac)=1-ab-bc-ac+a^2bc+ab^2c+abc^2-(abc)^2 \\\\\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-ab-bc-ac+abc(a+b+c)-(abc)^2 \\\\\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-ab-bc-ac+abc-(abc)^2 \\\\\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1-(1-a)(1-b)(1-c)-(abc)^2 \\\\\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ge 1-\frac 8{27}-\frac 1{729}=\frac{512}{729}=\left(\frac 89\right)^3\ (*) \)

Aplicam din nou \( AM-GM\ :\ \frac{\sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}}3\ \ge\ \sqrt[6]{(1-ab)(1-ac)(1-bc)}\ \ge^{(*)}\ \frac{2\sqrt 2}{3} \)

sau \( \sqrt {1 - bc} + \sqrt {1 - ac} + \sqrt {1 - ab}\geq 2\sqrt 2 \), c.c.t.d.