Question

Chứng minh:

a, \(\dfrac{a^2+3}{\sqrt{a^2+3}}>2\)

b,\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\) (a , b >0)

c,\(\left(\sqrt{a}+\sqrt{b}\right)^2\ge2\sqrt{2\left(a+b\right).\sqrt{ab}}\)

Answer 1

b)Áp dụng BĐT AM-GM ta có:

\(\dfrac{\sqrt{a}}{\sqrt{b}}+\dfrac{\sqrt{b}}{\sqrt{a}}\ge2\sqrt{\dfrac{\sqrt{a}}{\sqrt{b}}\cdot\dfrac{\sqrt{b}}{\sqrt{a}}}=2\)

Xảy ra khi \(a=b\)

c)Áp dụng BĐT \(x^2+y^2\ge2xy\) có:

\(VT=\left(\sqrt{a}+\sqrt{b}\right)^2=a+b+2\sqrt{ab}\)

\(\ge2\sqrt{\left(a+b\right)\cdot2\sqrt{ab}}=2\sqrt{2\left(a+b\right)\cdot\sqrt{ab}}=VP\)

Xảy ra khi \(a=b\)

Answer 2

a)\(\dfrac{a^2+3}{\sqrt{a^2+3}}=\sqrt{a^2+3}\ge\sqrt{3}< 2\)\

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