Cho $a,b,c>0$ thỏa mãn: $ab+bc+ca=\sqrt2$.
Chúng minh rằng: $\displaystyle \frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\le\frac{1}{\sqrt{abc}}$
Lời giải:
Áp dụng BĐT Bunhia ta có:
$\displaystyle \left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^2\le(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$
Ta chỉ cần chứng minh: $\displaystyle P=abc(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\le1$
Ta có:
$\displaystyle \frac{a^2bc(a+b+c)}{a+b}=a^2bc+\frac{a^2bc^2}{a+b}$
$\displaystyle \le a^2bc+\frac{a^2bc^2}{4}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\displaystyle =a^2bc+\frac{abc^2}{4}+\frac{a^2c^2}{4}$
Từ đó suy ra:
$\displaystyle P=abc(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$
$\displaystyle \le a^2bc+ab^2c+abc^2+\frac{a^2bc}{4}+\frac{ab^2c}{4}+\frac{abc^2}{4}+\frac{a^2b^2}{4}+\frac{b^2c^2}{4}+\frac{a^2c^2}{4}$
$\displaystyle \le a^2bc+ab^2c+abc^2+\frac{a^2b^2}{2}+\frac{b^2c^2}{2}+\frac{a^2c^2}{2}$
$\displaystyle =\frac{(ab+bc+ca)^2}{2}=1$, đpcm.
$\displaystyle \left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^2\le(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$
Ta chỉ cần chứng minh: $\displaystyle P=abc(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)\le1$
Ta có:
$\displaystyle \frac{a^2bc(a+b+c)}{a+b}=a^2bc+\frac{a^2bc^2}{a+b}$
$\displaystyle \le a^2bc+\frac{a^2bc^2}{4}\left(\frac{1}{a}+\frac{1}{b}\right)$
$\displaystyle =a^2bc+\frac{abc^2}{4}+\frac{a^2c^2}{4}$
Từ đó suy ra:
$\displaystyle P=abc(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$
$\displaystyle \le a^2bc+ab^2c+abc^2+\frac{a^2bc}{4}+\frac{ab^2c}{4}+\frac{abc^2}{4}+\frac{a^2b^2}{4}+\frac{b^2c^2}{4}+\frac{a^2c^2}{4}$
$\displaystyle \le a^2bc+ab^2c+abc^2+\frac{a^2b^2}{2}+\frac{b^2c^2}{2}+\frac{a^2c^2}{2}$
$\displaystyle =\frac{(ab+bc+ca)^2}{2}=1$, đpcm.
No comments:
Post a Comment