ta có \(\sum\) \(a+\frac{9}{16}a^2\ge\frac{3}{2}\sqrt{a^3}\)

\(\Rightarrow\)\(\sum\) \(a\ge\frac{3}{2}\sqrt{a^3}-\frac{9}{16}a^2\)\(\Rightarrow a+b+c\ge\frac{3}{2}(\sqrt{a^3}+\sqrt{b^3}+\sqrt{c^3})-\frac{9}{16}(a^2+b^2+c^2)\ge\frac{9}{2}\sqrt{abc}-\frac{9}{16}.4\sqrt{abc}\)>\(2\sqrt{abc}\) theo bđt côsi 

ĐPCM 

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a) \(x\ge0\)đặt \(\sqrt{x}=a\ge0\)

\(A=\frac{2a}{a^2-a+1}\Leftrightarrow A.a^2+A-2a=0\Leftrightarrow A.a^2-\left(A+2\right)a+A=0\)

\(\Delta=\left(A+2\right)^2-4A^2=-3A^2+4A+4\ge0\Rightarrow A\le2\)

\(\Rightarrow A_{max}=2\) khi  \(x=1\)

b) 

\(x\ge0\)

\(B=-\left(x-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)-\frac{7}{4}=-\left(\sqrt{x-\frac{1}{2}}\right)^2-\frac{7}{4}\le\frac{-7}{4}\)

\(\Rightarrow B_{max}=\frac{-7}{4}\) khi \(\sqrt{x=}\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)

c) \(x\ge0\)

\(C=-2+\sqrt{x}-1=-2\left(x-2.\sqrt{x}.\frac{1}{4}+\frac{1}{16}\right)-\frac{7}{8}\)

\(C=-2\left(\sqrt{x}-\frac{1}{4}\right)^2\frac{7}{8}\le\frac{-7}{8}\)

\(C_{max}=\frac{-7}{8}\)khi đó \(x=\frac{1}{16}\)

gt \(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=3\). Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\) suy ra \(xy+yz+zx=3\)

Quy về:Tìm Min \(A=\Sigma_{cyc}\frac{x^3}{\left(2z+y\right)}\)

2+3=5

Mộc Tử Hạ

k6,nhớ add

5

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