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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(T=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)
\(\geq \frac{1}{2}.3\sqrt[3]{\frac{1}{abc}}=\frac{3}{2}\) (theo BĐT AM-GM)
Vậy $T_{\min}=\frac{3}{2}$.
Giá trị này đạt tại $a=b=c=1$
Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=1\)
BĐT trở thành: \(\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}+\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}+\dfrac{zx}{\sqrt{x^2+z^2+2y^2}}\le\dfrac{1}{2}\)
Ta có:
\(x^2+z^2+y^2+z^2\ge\dfrac{1}{2}\left(x+z\right)^2+\dfrac{1}{2}\left(y+z\right)^2\ge\left(x+z\right)\left(y+z\right)\)
\(\Rightarrow\dfrac{xy}{\sqrt{x^2+y^2+2z^2}}\le\dfrac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{xy}{x+z}+\dfrac{xy}{y+z}\right)\)
Tương tự: \(\dfrac{yz}{\sqrt{y^2+z^2+2x^2}}\le\dfrac{1}{2}\left(\dfrac{yz}{x+y}+\dfrac{yz}{x+z}\right)\)
\(\dfrac{zx}{\sqrt{z^2+x^2+2y^2}}\le\dfrac{1}{2}\left(\dfrac{zx}{x+y}+\dfrac{zx}{y+z}\right)\)
Cộng vế với vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{zx+yz}{x+y}+\dfrac{xy+zx}{y+z}+\dfrac{yz+xy}{z+x}\right)=\dfrac{1}{2}\left(x+y+z\right)=\dfrac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c\)
$\sum \sqrt{\frac{ab+2c^2}{1+ab-c^2}}\geq ab+bc+ca+2$ - Bất đẳng thức và cực trị - Diễn đàn Toán học
Lời giải:
Điều kiện \(ab+bc+ac=abc\) là không cần thiết và bạn cần sửa lại đề bài là: CMR \(\sqrt{\frac{b^2+2a^2}{ab}}+\sqrt{\frac{c^2+2b^2}{bc}}+\sqrt{\frac{a^2+2c^2}{ac}}\geq 3\sqrt{3}\)
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Áp dụng BĐT AM-GM ta có:
\(b^2+2a^2=b^2+a^2+a^2\geq 3\sqrt[3]{b^2a^4}\)
\(\Rightarrow \frac{b^2+2a^2}{ab}\geq \frac{3\sqrt[3]{b^2a^4}}{ab}=3\sqrt[3]{\frac{a}{b}}\)
\(\Rightarrow \sqrt{\frac{b^2+2a^2}{ab}}\geq \sqrt{3}.\sqrt[6]{\frac{a}{b}}\)
Hoàn toàn TT: \(\sqrt{\frac{c^2+2b^2}{bc}}\geq \sqrt{3}.\sqrt[6]{\frac{b}{c}}; \sqrt{\frac{a^2+2c^2}{ac}}\geq \sqrt{3}.\sqrt[6]{\frac{c}{a}}\)
Cộng theo vế những BĐT vừa thu được:
\(\Rightarrow \text{VT}\geq \sqrt{3}\left(\sqrt[6]{\frac{a}{b}}+\sqrt[6]{\frac{b}{c}}+\sqrt[6]{\frac{c}{a}}\right)\)
\(\geq \sqrt{3}.3\sqrt[18]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}=3\sqrt{3}\) (tiếp tục áp dụng BĐT AM-GM)
Vậy ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{1-a-b-ab}}=\sqrt{\dfrac{ab}{\left(1-b\right)\left(1-a\right)}}\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}}{2}\left(1\right)\) \(tương-tự\Rightarrow\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{\dfrac{b}{1-c}+\dfrac{c}{1-b}}{2}\left(2\right)\)
\(\Rightarrow\sqrt{\dfrac{ca}{b+ ca}}\le\dfrac{\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}\left(3\right)\)
\( \left(1\right)\left(2\right)\left(3\right)\Rightarrow A\le\dfrac{\dfrac{a}{1-b}+\dfrac{b}{1-a}+\dfrac{b}{1-c}+\dfrac{c}{1-b}+\dfrac{c}{1-a}+\dfrac{a}{1-c}}{2}=\dfrac{\dfrac{a+c}{1-b}+\dfrac{b+c}{1-a}+\dfrac{b+a}{1-c}}{2}=\dfrac{\dfrac{1-b}{1-b}+\dfrac{1-a}{1-a}+\dfrac{1-c}{1-c}}{2}=\dfrac{3}{2}\)
\(\Rightarrow A_{max}=\dfrac{3}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Áp dụng BĐT AM-GM ta có:
\(P=\sqrt{\dfrac{2a}{2b+2c-a}}+\sqrt{\dfrac{2b}{2c+2a-b}}+\sqrt{\dfrac{2c}{2a+2b-c}}\)
\(=\dfrac{\sqrt{6}a}{\sqrt{3a\left(2b+2c-a\right)}}+\dfrac{\sqrt{6}b}{\sqrt{3b\left(2c+2a-b\right)}}+\dfrac{\sqrt{6}c}{\sqrt{3c\left(2a+2b-c\right)}}\)
\(\ge\dfrac{\sqrt{6}a}{\dfrac{3a+2b+2c-a}{2}}+\dfrac{\sqrt{6}b}{\dfrac{3b+2c+2a-b}{2}}+\dfrac{\sqrt{6}c}{\dfrac{3c+2a+2b-c}{2}}\)
\(\ge\dfrac{\sqrt{6}a}{a+b+c}+\dfrac{\sqrt{6}b}{a+b+c}+\dfrac{\sqrt{6}c}{a+b+c}\)
\(=\dfrac{\sqrt{6}\left(a+b+c\right)}{a+b+c}=\sqrt{6}\)
\(\sum\dfrac{ab}{\sqrt{c+ab}}=\sum\dfrac{ab}{\sqrt{c\left(a+b+c\right)+ab}}=\sum\dfrac{ab}{\sqrt{\left(c+a\right)\left(c+b\right)}}\le\dfrac{1}{2}\sum\left(\dfrac{ab}{a+b}+\dfrac{ab}{a+c}\right)=\dfrac{a+b+c}{2}=\dfrac{1}{2}\)
GTNN của P là \(\dfrac{1}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Ta có: \(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{5}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2}\ge\sqrt{\dfrac{5}{4}}\left(a+b\right)\)
Cmtt ta có: \(\sqrt{2b^2+bc+2c^2}\ge\sqrt{\dfrac{5}{4}}\left(b+c\right)\)
\(\sqrt{2c^2+ca+2a^2}\ge\sqrt{\dfrac{5}{4}}\left(c+a\right)\)
\(\Rightarrow P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\dfrac{\sqrt{5}}{3}\)
Dấu "=" xảy ra <=> a = b = c = \(\dfrac{1}{9}\)
