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Lời giải:
Đặt biểu thức đã cho là \(A\)
Ta có:
\(6a^2+8ab+11b^2=2a^2+(2a+2b)^2+7b^2\)
Áp dụng BĐT Bunhiacopxky:
\([2a^2+(2a+2b)^2+7b^2](2+4^2+7)\geq (2a+8a+8b+7b)^2\)
\(\Leftrightarrow 25(6a^2+8ab+11b^2)\geq (10a+15b)^2\)
\(\Rightarrow \sqrt{6a^2+8ab+11b^2}\geq 2a+3b\)
\(\Rightarrow \frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\leq \frac{a^2+3ab+b^2}{2a+3b}\)
Thực hiện tương tự với các biểu thức còn lại và cộng theo vế:
\(A\leq \frac{a^2+3ab+b^2}{2a+3b}+\frac{a^2+3ac+c^2}{2c+3a}+\frac{b^2+3bc+c^2}{2b+3c}\)
\(6A\leq \frac{3a(2a+3b)+2b(2a+3b)+5ab}{2a+3b}+\frac{3c(2c+3a)+2a(2c+3a)+5ac}{2c+3a}+\frac{3b(2b+3c)+2c(2b+3c)+5bc}{2b+3c}\)
\(\Leftrightarrow 6A\leq 3a+2b+\frac{5ab}{2a+3b}+3c+2a+\frac{5ac}{2c+3a}+3b+2c+\frac{5bc}{2b+3c}\)
\(\Leftrightarrow 6A\leq 5(a+b+c)+5\left(\frac{ab}{2a+3b}+\frac{bc}{2b+3c}+\frac{ac}{2c+3a}\right)\)
Theo hệ quả của BĐT AM-GM:
\((a+b+c)^2\leq 3(a^2+b^2+c^2)=9\Rightarrow a+b+c\leq 3(1)\)
Áp dụng BĐT Cauchy-Schwarz dạng ngược:
\(\frac{ab}{2a+3b}\leq \frac{ab}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}\right)\)
\(\frac{bc}{2b+3c}\leq \frac{bc}{25}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{ca}{2c+3a}\leq \frac{ca}{25}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}\right)\)
\(\Rightarrow \frac{ab}{2a+3b}+\frac{bc}{2b+3c}+\frac{ac}{2c+3a}\leq \frac{1}{5}(a+b+c)(2)\)
Từ (1); (2) suy ra:
\(6A\leq 5(a+b+c)+5.\frac{1}{5}(a+b+c)=6(a+b+c)\leq 18\)
\(\Rightarrow A\leq 3\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
\(P=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\Leftrightarrow\sqrt{\dfrac{97}{4}}P=\sqrt{4+\dfrac{81}{4}}\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{4+\dfrac{81}{4}}\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\ge\left(2a+\dfrac{9}{2a}\right)+\left(2b+\dfrac{9}{2b}\right)+\left(2c+\dfrac{9}{2c}\right)\)
\(=2\left(a+b+c\right)+\dfrac{9}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\ge4+\dfrac{9}{2}.\dfrac{9}{a+b+c}=4+\dfrac{81}{4}=\dfrac{97}{4}\)
\(\Rightarrow P\ge\sqrt{\dfrac{97}{4}}\)
PS: Lần sau chép đề cẩn thận nhé bạn.
Nếu là \(\ge \) thì easy rồi. Áp dụng BĐT Min....
\(VT=\sqrt{a^2+\dfrac{1}{a^2}}+\sqrt{b^2+\dfrac{1}{b^2}}+\sqrt{c^2+\dfrac{1}{c^2}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{9}{a+b+c}\right)^2}\)
\(\ge\sqrt{2^2+\left(\dfrac{9}{2}\right)^2}=\sqrt{\dfrac{97}{4}}=VP\)
Khi \(a=b=c=\frac{2}{3}\)
Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)
Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)
\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)
\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)
Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)
\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)
\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\left\{{}\begin{matrix}3+a^2\ge\left(a+c\right)\left(a+b\right)\\3+b^2\ge\left(a+b\right)\left(b+c\right)\\3+c^2\ge\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{\sqrt{3+a^2}}\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\dfrac{ca}{\sqrt{3+b^2}}\le\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\\\dfrac{ab}{\sqrt{3+c^2}}\le\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(\Leftrightarrow VT\le\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{bc}{a+c}+\dfrac{bc}{a+b}}{2}\\\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{ab}{a+c}+\dfrac{ab}{b+c}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)}{2}\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\) (2)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (3)
Từ (1) , (2) , (3)
\(\Rightarrow VT\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ca}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=1\)
Ta luôn có \(\left(\dfrac{1}{\sqrt{a}}-\dfrac{1}{\sqrt{b}}\right)^2\ge0\forall a;b\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\)
\(\Leftrightarrow2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{2}{\sqrt{ab}}+\dfrac{1}{a}+\dfrac{1}{b}\)
\(\Leftrightarrow\dfrac{2\left(a+b\right)}{ab}\ge\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)^2\)
\(\Leftrightarrow\sqrt{\dfrac{2\left(a+b\right)}{ab}}\ge\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(\sqrt{2}\left(\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\right)\ge2\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)
\(\Leftrightarrow\sqrt{\dfrac{a+b}{ab}}+\sqrt{\dfrac{b+c}{bc}}+\sqrt{\dfrac{a+c}{ac}}\ge\sqrt{\dfrac{2}{a}}+\sqrt{\dfrac{2}{b}}+\sqrt{\dfrac{2}{c}}\)
\("="\Leftrightarrow a=b=c\)