M= ( x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+ x^2 với x= 1/2a +1/2b+1/2c
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\(A=\dfrac{x-4+5}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{5}{\sqrt{x}-2}\)
\(=\sqrt{x}-2+\dfrac{5}{\sqrt{x}-2}+4\ge2\sqrt{\dfrac{5\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+4=4+2\sqrt{5}\)
\(A_{min}=4+2\sqrt{5}\) khi \(9+4\sqrt{5}\)
b.
Đặt \(\left(a;b;c\right)=\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{l}{z}\right)\Rightarrow xyz=1\)
\(B=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)
\(B_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\Rightarrow a=b=c=1\)
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Từ \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c=\frac{1}{2}.\left(a+b+c\right)\Rightarrow2x=a+b+c\)
\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)
\(=x^2-xb-ax+ab+x^2-xc-bx+bc+x^2-ax-cx+ac+x^2\)
\(=4x^2-2ax-2bx-2cx+ab+bc+ac\)
\(=4x^2-2x\left(a+b+c\right)+ab+bc+ca\)
Thay 2x=a+b+c,ta đc:
\(M=4x^2-2x.2x+ab+bc+ca=4x^2-4x^2+ab+bc+ca=ab+bc+ca\)
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Bài 2:
Từ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\rightarrow\left(x;y;z\right)\Rightarrow\hept{\begin{cases}x,y,z>0\\x+y+z=2\end{cases}}\)
\(BDT\Leftrightarrow\frac{x^3}{\left(x-2\right)^2}+\frac{y^3}{\left(y-2\right)^2}+\frac{z^3}{\left(z-2\right)^2}\ge\frac{1}{2}\)
Ta chứng minh bổ đề \(\frac{x^3}{\left(x-2\right)^2}\ge x-\frac{1}{2}\Leftrightarrow\frac{\left(3x-2\right)^2}{\left(x-2\right)^2}\ge0\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{y^3}{\left(y-2\right)^2}\ge y-\frac{1}{2};\frac{z^3}{\left(z-2\right)^2}\ge z-\frac{1}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\left(x+y+z\right)-\frac{3}{2}=2-\frac{3}{2}=\frac{1}{2}=VP\)