giải hệ \(\hept{\begin{cases}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\\x^2y+8\sqrt{y-2}=35\end{cases}}\)
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\(P=\frac{3x-6\sqrt{x}+7}{2\sqrt{x}-2}+\frac{y-4\sqrt{x}+10}{\sqrt{y}-2}\)
\(=\frac{3\left(\sqrt{x}-1\right)}{2}+\frac{4}{2\left(\sqrt{x}-1\right)}+\left(\sqrt{y}-2\right)+\frac{6}{\sqrt{y-1}}\)
\(=\frac{3\left(\sqrt{x}-1\right)}{2}+\frac{3}{2\left(\sqrt{x}-1\right)}+\left(\sqrt{y}-2\right)+\frac{4}{\left(\sqrt{y}-2\right)}+\frac{4}{2\left(\sqrt{y}-2\right)}+\frac{1}{2\left(\sqrt{x}-1\right)}\)
\(\ge2.\sqrt{\frac{3}{2}.\frac{3}{2}}+2\sqrt{4}+\frac{\left(1+2\right)^2}{2\left(\sqrt{x}+\sqrt{y}-3\right)}\)
\(=3+4+\frac{3}{2}=\frac{17}{2}\)
Dấu "=" xảy ra <=> x = 4 và y = 16
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Ta phải chứng minh
\(\displaystyle \sum\)\(\frac{1+a}{b+c}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\)\(\displaystyle \sum\)\(\frac{2a+b+c}{b+c}\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\)\(\displaystyle \sum\)\(\frac{2a}{b+c}+3\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)
\(\Leftrightarrow\frac{a}{b}-\frac{a}{b+c}+\frac{b}{c}-\frac{b}{b+c}+\frac{c}{a}-\frac{c}{a+b}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{bc}{a\left(a+b\right)}+\frac{ab}{c\left(c+a\right)}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\left(ac\right)^2}{abc\left(b+c\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}+\frac{\left(ca\right)^2}{abc\left(c+a\right)}\ge\frac{3}{2}\)
Mặt khác: Theo BĐT AM-GM ta có:
\(\left(ab+bc+ca\right)^2\ge3\left(a^2bc+ab^2c+abc^2\right)=3abc\left(a+b+c\right)\)
Theo BĐT Cauchy-Schwwarz ta có:
\(\frac{\left(ac\right)^2}{abc\left(a+b\right)}+\frac{\left(bc\right)^2}{abc\left(a+b\right)}+\frac{\left(ca\right)^2}{abc\left(c+a\right)}\ge\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\frac{3}{2}\)
Bài toán được chứng minh xong. Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
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