Question

cho x,y \(\ge\)0 và \(x^2+y^2=1.T\text{ín}h\) GTNN của P = \(\sqrt{1+2x}+\sqrt{1+2y}\)

Answer 1

Ta có \(x,y\le1\) nên \(1\le\sqrt{1+2x}\le\sqrt{3}\).

Suy ra \(\left(\sqrt{1+2x}-1\right)\left(\sqrt{1+2x}-\sqrt{3}\right)\le0\Rightarrow\left(\sqrt{3}+1\right)\sqrt{1+2x}\ge1+2x+\sqrt{3}\).

Tương tự \(\left(\sqrt{3}+1\right)\sqrt{1+2y}\ge1+2y+\sqrt{3}\).

Suy ra \(\left(\sqrt{3}+1\right)P\ge2+2\sqrt{3}+2\left(x+y\right)\).

Mà \(\left(x+y\right)^2\ge x^2+y^2=1\Rightarrow x+y\le1\Rightarrow\left(\sqrt{3}+1\right)P\ge2+2\sqrt{3}+2=4+2\sqrt{3}\Rightarrow P\ge\sqrt{3}+1\).

Dấu "=" xảy ra khi x = 0; y = 1 hoặc x = 1; y = 0.

Answer 2

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