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\(\left(2\sqrt{1+a}\right)^2=4\left(1+a\right)=\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)
\(\Leftrightarrow\)\(x+y\ge2a\)
\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow1\ge4xy\Leftrightarrow xy\le\frac{1}{4}\)(1)
\(\left(x-y\right)^2\ge0\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge2\Leftrightarrow x+y\ge\sqrt{2}\)
Từ phần a ta có \(x+y\le\sqrt{2}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(VT^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2\)
\(\le\left(1+1\right)\left(2\left(x+y\right)+2\right)\)
\(=2\cdot\left(2\left(x+y\right)+2\right)\le2\cdot\left(2\sqrt{2}+2\right)\)
\(=4\sqrt{2}+4=VP^2\)
Suy ra \(VT\ge VP\) (ĐPCM)
Áp dụng bất đẳng thức Schwartz , ta có :
\(\left(1.\sqrt{1+x}+1.\sqrt{1+y}\right)^2\le\left(1^2+1^2\right)\left(1+x+1+y\right)\)
\(\Leftrightarrow4\left(1+a\right)\le2.\left(x+y+2\right)\)
\(\Leftrightarrow x+y+2\ge2a+2\)
\(\Rightarrow x+y\ge2a\left(ĐPCM\right)\)
a, \(\left(\sqrt{3}-\sqrt{2}\right)\cdot\sqrt{5+2\sqrt{6}}=\sqrt{15+2\cdot3\cdot\sqrt{6}}-\sqrt{10+2\cdot2\cdot\sqrt{6}}=\sqrt{9+2\cdot3\cdot\sqrt{6}+6}-\sqrt{6+2\cdot\sqrt{6}\cdot2+4}=\sqrt{\left(3+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}+2\right)^2}=3+\sqrt{6}-\sqrt{6}-2=3-2=1\left(đpcm\right)\)
b, đề không rõ ràng
BĐT C-S:
\(\left(2\sqrt{a+1}\right)^2=\left(\sqrt{x+1}+\sqrt{y+1}\right)^2\)
\(\le\left(1+1\right)\left(x+1+y+1\right)=2\left(x+y+2\right)\)
Hay \(4\left(a+1\right)\le2\left(x+y+2\right)\)
\(\Leftrightarrow2a+2\le x+y+2\Leftrightarrow2a\le x+y\) *DDungs*