CMR: Với mọi x,y thì \(5x^2+3y^2+5xy-x-3y+1\geq0\)
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Bunyakovsky nhé :)
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Còn không thì Cauchy
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
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\(ĐK:\frac{2}{3}\ge x\ge\frac{5}{2}\)
\(PT\Leftrightarrow\left(4x^2-4x+1\right)+\left(2x-5\right)\sqrt{2+4x}-\left(2x+3\right)\sqrt{6-4x}+16=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\sqrt{2+4x}-\left(2x+3\right)\sqrt{6-4x}+16=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\left(\sqrt{2+4x}-2\right)-\left(2x+3\right)\left(\sqrt{6-4x}-2\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\frac{2+4x-4}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{6-4x-4}{\sqrt{6-4x}+2}=0\)
\(\Leftrightarrow\left(2x-1\right)^2+\left(2x-5\right)\frac{2\left(2x-1\right)}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2\left(2x-1\right)}{\sqrt{6-4x}+2}=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x-1+\left(2x-5\right)\frac{2}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2}{\sqrt{6-4x}+2}\right)=0\)
Theo ĐK ta chứng minh đc \(\left(2x-1+\left(2x-5\right)\frac{2}{\sqrt{2+4x}+2}+\left(2x+3\right)\frac{-2}{\sqrt{6-4x}+2}\right)>0\)
Do đó \(2x-1=0\Rightarrow x=\frac{1}{2}\left(TMĐKXĐ\right)\)
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HPT \(\Leftrightarrow\hept{\begin{cases}x=2\\2m+y=m^2+3\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=m^2-2m+3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\x+y=m^2-2m+5\end{cases}}}\)
Ta có : \(x+y=m^2-2m+5=\left(m-1\right)^2+4\ge4\forall m\)
Dấu "=" xảy ra \(\Leftrightarrow m=1\)
Vậy \(x+y\) đạt GTNN là 4 tại \(m=1\)