Question

[Nguyễn Minh Tuấn]

Bài tập vận dụng: Cho các số thực dương CMR:

\(a.\frac{{{x^2}}}{{{x^2} + 2yz}} + \frac{{{y^2}}}{{{y^2} + 2zx}} + \frac{{{z^2}}}{{{z^2} + 2xy}} \ge 1\)

\(b.\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} \ge \frac{3}{2}\)

Answer 1

[Lam Ngo Tung]

a) Vì x;y;z > 0 nên áp dụng bất đẳng thức Bunhiakovsky : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) , ta được :

\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}\)

\(\Leftrightarrow\)\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Vậy \(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\ge1\left(ĐPCM\right)\)

b) Ta chứng minh bất đẳng thức phụ :\(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)

\(\Leftrightarrow\left(a+b+c\right)^2-3\left(ab+bc+ac\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3ac-3bc\ge0\)

\(\Leftrightarrow a^2+b^2+c^2-ab-ab-ac\ge0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) ( luôn đúng )

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+ab+ac\right)\)

Vì a,b,c > 0 nên áp dụng bất đẳng thức Bunhiakovsky : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) , ta được :

\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\)

mà \(\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)

Vậy \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\left(ĐPCM\right)\)