Áp dụng BĐT Cô si ta có \(x+\dfrac{4}{x}\ge2\sqrt{x.\dfrac{4}{x}}=2\sqrt{4}=4\)

Dấu = xảy ra \(\Leftrightarrow x=\dfrac{4}{x}\Leftrightarrow x^2=4\Leftrightarrow x=2\) (Vì \(1\le x\le3\))

\(y=\frac{5x}{x^2+4}\le\frac{5x}{2\sqrt{x^2.4}}=\frac{5}{4}\)

Dấu "=" xảy ra khi \(x=2\)

\(y=\frac{x^2}{\left(x^2+\frac{3}{2}+\frac{3}{2}\right)^3}\le\frac{x^2}{\left(3\sqrt[3]{x^2.\frac{3}{2}.\frac{3}{2}}\right)^3}=\frac{4x^2}{243x^2}=\frac{4}{243}\)

Dấu "=" xảy ra khi \(x=\frac{\sqrt{6}}{2}\)

1 ) Áp dụng BĐT Cô - si cho a ; b dương , ta có :

\(a+b\ge2\sqrt{ab}\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\left(đpcm\right)\)

2 ) \(\dfrac{2}{xy}+\dfrac{3}{x^2+y^2}=\dfrac{3}{2xy}+\dfrac{3}{x^2+y^2}+\dfrac{1}{2xy}=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\)

\(\ge3.\dfrac{4}{\left(x+y\right)^2}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{3.4}{1}+\dfrac{1}{\dfrac{1}{2}}=12+2=14\)

( áp dụng BĐT Cô - si cho 2 số x ; y dương và BĐT phụ \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Vậy ...

Ta có \(\dfrac{1}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{\dfrac{1}{\sqrt{x}}.\sqrt{x}}=2;\dfrac{1}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{\dfrac{1}{\sqrt{y}}.\sqrt{y}}=2\)

=> VT\(\ge4\)

dấu = xảy ra <=> x=y=1 (thỏa mãn điều kiện )

\(A=\frac{\sqrt[4]{3}}{2}.\frac{2x}{\sqrt[4]{3}}\sqrt{4-x^4}\le\frac{\sqrt[4]{3}}{4}\left(\frac{4x^2}{\sqrt{3}}+4-x^4\right)=\frac{\sqrt[4]{3}}{4}\left[\frac{16}{3}-\left(x^2-\frac{2\sqrt{3}}{3}\right)^2\right]\le\frac{4\sqrt[4]{3}}{3}\)

\(A_{max}=\frac{4\sqrt[4]{3}}{3}\) khi \(x^2=\frac{2\sqrt{3}}{3}\)