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Đặt \(\hept{\begin{cases}\sqrt{a^2+b^2}=x\\\sqrt{b^2+c^2}=y\\\sqrt{c^2+a^2}=z\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=1\end{cases}}\)
Và \(\hept{\begin{cases}a^2=\frac{x^2+z^2-y^2}{2}\\b^2=\frac{x^2+y^2-z^2}{2}\\c^2=\frac{y^2+z^2-x^2}{2}\end{cases}}\) và \(\hept{\begin{cases}b+c\le\sqrt{2\left(b^2+c^2\right)}=\sqrt{2}y\\a+b\le\sqrt{2}x\\c+a\le\sqrt{2}z\end{cases}}\)
\(\Rightarrow VT\ge\frac{1}{2\sqrt{2}}\left(\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{2z}+\frac{y^2+z^2-x^2}{x}\right)\)
\(\ge\frac{1}{2\sqrt{2}}\left(\frac{2\left(x+y+z\right)^2}{x+y+z}-\left(x+y+z\right)\right)\)
\(=\frac{1}{2\sqrt{2}}\left(x+y+z\right)=\frac{1}{2\sqrt{2}}\)
\(a^2\sqrt{a}+b^2\sqrt{b}+c^2\sqrt{c}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)
\(=\left(a^2\sqrt{a}+\frac{1}{\sqrt{a}}\right)+\left(b^2\sqrt{b}+\frac{1}{\sqrt{b}}\right)+\left(c^2\sqrt{c}+\frac{1}{\sqrt{c}}\right)\)
\(\ge2a+2b+2c\ge6\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=6\)
Bài này ngược dấu hay sao ý:
Ta dự đoán dấu "=" xảy ra tại a = b = c =1
Áp dụng BĐT Cauchy-Schwarz: \(VT=\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\ge\frac{9}{a+2b}\) (1)
Ta có: \(a^2+1\ge2a;2b^2+2\ge4b\Rightarrow a^2+2b^2+3=3c^2+3\ge2\left(a+2b\right)\)
\(\Rightarrow\frac{3c^2+3}{2}\ge a+2b\).Suy ra:\(\frac{9}{a+2b}\ge\frac{18}{3c^2+3}=\frac{6}{c^2+1}\) (2)
Ta sẽ c/m: \(\frac{6}{c^2+1}\ge\frac{3}{c}\).Ta có: \(VT=\frac{6}{c^2+1}=6\left(1-\frac{c^2}{c^2+1}\right)=6-\frac{6c^2}{c^2+1}\ge6-\frac{6c^2}{2c}=6-3c\) (3)
Ta sẽ c/m: \(6-3c\ge\frac{3}{c}\Leftrightarrow3c+\frac{3}{c}\le6\).Mặt khác,theo AM-GM
\(3c+\frac{3}{c}\ge2.\sqrt{3c.\frac{3}{c}}=2.3=6\Rightarrow\) mâu thuẫn?
với mọi x,y,z >0 ta có: \(x+y+z\ge3\sqrt[3]{xyz};\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{xyz}}\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\ge9\)
\(\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
đẳng thức xảy ra khi x=y=z
ta có: \(5a^2+2ab+2b^2=\left(2a+b\right)^2+\left(a-b\right)^2\ge\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}\right)\)
đẳng thức xảy ra khi a=b
tương tự: \(\frac{1}{\sqrt{5b^2+2ab+2b^2}}\le\frac{1}{2b+c}\le\frac{1}{9}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
đẳng thức xảy ra khi b=c
\(\frac{1}{\sqrt{5c^2+2bc+2c^2}}\le\frac{1}{2c+a}\le\frac{1}{9}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)
đẳng thức xảy ra khi c=a
Vậy \(\frac{1}{\sqrt{5a^2+2ca+2a^2}}+\frac{1}{\sqrt{5b^2+2bc+2c^2}}+\frac{1}{\sqrt{5c^2+2ac+2a^2}}\le\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{2}{3}\)
đẳng thức xảy ra khi a=b=c=\(\frac{3}{2}\)
Đặt \(m=a^2+b^2+c^2,m\ge0\)
Áp dụng bất đẳng thức Bunhiacopxki , ta có :
\(\frac{9}{4}=\left(a.\sqrt{1-b^2}+b.\sqrt{1-c^2}+c.\sqrt{1-a^2}\right)^2\le\left(a^2+b^2+c^2\right)\left(3-a^2-b^2-c^2\right)\)
\(\Rightarrow m\left(3-m\right)\ge\frac{9}{4}\) \(\Leftrightarrow\left(m-\frac{3}{2}\right)^2\le0\) mà ta luôn có \(\left(m-\frac{3}{2}\right)^2\ge0\)
Do đó \(\left(m-\frac{3}{2}\right)^2=0\Rightarrow m=\frac{3}{2}\)
Vậy \(a^2+b^2+c^2=\frac{3}{2}\)
Ta có :\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Leftrightarrow\frac{a+b}{ab}=\frac{1}{c}\Leftrightarrow c=\frac{ab}{a+b}\)
\(\Rightarrow c^2=\frac{a^2b^2}{\left(a+b\right)^2}\) thay vào A ta được :
\(A=\sqrt{a^2+b^2+\frac{a^2b^2}{\left(a+b\right)^2}}=\sqrt{\frac{a^2\left(a+b\right)^2+b^2\left(a+b\right)^2+a^2b^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\frac{a^4+2a^3b+a^2b^2+a^2b^2+2ab^3+b^4+a^2b^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\frac{a^4+b^4+a^2b^2+2a^3b+2ab^3+2a^2b^2}{\left(a+b\right)^2}}\)
\(=\sqrt{\frac{\left(a^2+b^2+ab\right)^2}{\left(a+b\right)^2}}=\frac{a^2+b^2+ab}{a+b}\in Q\forall a;b;c\in Q^+\)(dpcm)