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cho a,b,c > 0 . Tìm GTNN của
\(P=\frac{2+a^3}{2+a+b^3}+\frac{2+b^3}{2+b+c^3}+\frac{2+c^3}{2+c+a^3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow17\cdot\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)
\(\Leftrightarrow\sqrt{17}\cdot\sqrt{a^2+\frac{1}{b^2}}\ge a+\frac{4}{b}\)
Tương tự ta có :
\(\sqrt{17}\cdot\sqrt{b^2+\frac{1}{c^2}}\ge b+\frac{4}{c}\)
\(\sqrt{17}\cdot\sqrt{c^2+\frac{1}{a^2}}\ge c+\frac{4}{a}\)
Cộng theo vế của 3 bđt ta được :
\(\sqrt{17}\cdot\left(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\right)\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(\Leftrightarrow\sqrt{17}\cdot A\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
Áp dụng bất đẳng thức Cô-si :
\(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)
\(=16a+\frac{4}{a}+16b+\frac{4}{b}+16c+\frac{4}{c}-15a-15b-15c\)
\(\ge2\sqrt{\frac{4\cdot16a}{a}}+2\sqrt{\frac{4\cdot16b}{b}}+2\sqrt{\frac{4\cdot16c}{c}}-15\left(a+b+c\right)\)
\(\ge16+16+16-15\cdot\frac{3}{2}=\frac{51}{2}\)
Do đó : \(\sqrt{17}\cdot A\ge\frac{51}{2}\)
\(\Leftrightarrow A\ge\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a^3}{\left(1-a\right)^2}+\frac{1-a}{8}+\frac{1-a}{8}\ge3\sqrt[3]{\frac{a^3}{\left(1-a\right)^2}.\frac{\left(1-a\right)}{8}.\frac{1-a}{8}}=\frac{3a}{4}\)
Suy ra \(\frac{a^3}{1-a^2}\ge\frac{3a}{4}-\frac{\left(1-a\right)}{4}=\frac{4a-1}{4}\)
Tương tự hai BĐT còn lại rồi cộng theo vế:
\(A\ge\frac{4\left(a+b+c\right)-3}{4}=\frac{1}{4}\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có:
$a^2+b^2+c^2+ab+bc+ac=\frac{6(a^2+b^2+c^2+ab+bc+ac)}{6}=\frac{4(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2}{6}$
$\geq \frac{(a-b)^2+(b-c)^2+(c-a)^2}{6}$
$\Rightarrow P\geq \frac{(a-b)^2+(b-c)^2+(c-a)^2}{6}.\left[\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\right]$
Đặt $a-b=m, b-c=n$ thì $a-c=m+n$
Khi đó:
$6P\geq [m^2+n^2+(m+n)^2]\left[\frac{1}{m^2}+\frac{1}{n^2}+\frac{1}{(m+n)^2}\right]$
Áp dụng BĐT AM-GM và Cauchy-Schwarz:
$[m^2+n^2+(m+n)^2]\left[\frac{1}{m^2}+\frac{1}{n^2}+\frac{1}{(m+n)^2}\right]$
$\geq [\frac{(m+n)^2}{2}+(m+n)^2]\left[\frac{1}{2}(\frac{1}{m}+\frac{1}{n})^2+\frac{1}{(m+n)^2}\right]$
$\geq \frac{3}{2}.(m+n)^2\left[\frac{8}{(m+n)^2}+\frac{1}{(m+n)^2}\right]$
$=\frac{3}{2}(m+n)^2.\frac{9}{(m+n)^2}=\frac{27}{2}$
$\Rightarrow 6P\geq \frac{27}{2}$
$\Rightarrow P\geq \frac{9}{4}$
Vậy GTNN của $P$ là $\frac{9}{4}$.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{a}{1+b^2}=\frac{a.\left(1+b^2\right)-ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\)
Mặt khác có : \(1+b^2\ge2b\Rightarrow\frac{ab^2}{1+b^2}\le\frac{ab^2}{2b}=\frac{ab}{2}\)
\(\Rightarrow-\frac{ab^2}{1+b^2}\ge-\frac{ab}{2}\Rightarrow a-\frac{ab^2}{1+b^2}\ge a-\frac{ab}{2}\)
Thiết lập tương tự với các phân thức còn lại ta có :
\(P\ge a+b+c-\frac{ab+bc+ca}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=3-\frac{3}{2}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow a=b=c=1\)
\(A=\frac{a\left(a+b^2\right)-ab^2}{a+b^2}+\frac{b\left(b+c^2\right)-bc^2}{b+c^2}+\frac{c\left(c+a^2\right)-ca^2}{c+a^2}\)
\(A=a+b+c-\left(\frac{ab^2}{a+b^2}+\frac{bc^2}{b+c^2}+\frac{ca^2}{c+a^2}\right)\)
\(A\ge3-\left(\frac{ab^2}{2\sqrt{ab^2}}+\frac{bc^2}{2\sqrt{bc^2}}+\frac{ca^2}{2\sqrt{ca^2}}\right)\)
\(A\ge3-\frac{1}{2}\left(\sqrt{a}.b+\sqrt{b}.c+\sqrt{c}.a\right)\)
\(A\ge3-\frac{1}{2}\left(\frac{b\left(a+1\right)}{2}+\frac{c\left(b+1\right)}{2}+\frac{a\left(c+1\right)}{2}\right)\)
\(A\ge3-\frac{1}{4}\left(ab+bc+ca+a+b+c\right)\)
\(A\ge3-\frac{1}{4}\left(3+\frac{\left(a+b+c\right)^2}{3}\right)\) \(=3-\frac{3}{2}=\frac{3}{2}\)
Min \(A=\frac{3}{2}\) \(\Leftrightarrow a=b=c=1\)