Tham khảo:

đề cho thêm x nữa hén=) , p/s đưa đề đàng hoàng có thịn cảm ng làm hén , ụa mà hiện hồn là sao.-. ghét nghỉ=))

ủa cái z dị pa :)

ÁP dụng AM-GM:

\(\sum\dfrac{a^2}{\sqrt{1-a^2}}=\sum\dfrac{a^3}{\sqrt{\left(1-a^2\right).a^2}}\ge\sum\dfrac{a^3}{\dfrac{1}{2}\left(1-a^2+a^2\right)}=2\sum a^3=2\left(đpcm\right)\)

Dấu = không xảy ra

Xài Bunhiacopxki thì bài này sẽ hơi dài:

Đặt vế trái là P

Ta có:

\(\left(\dfrac{1}{4}+4\right)\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)

\(\Leftrightarrow\dfrac{17}{4}\left(x^2+\dfrac{1}{x^2}\right)\ge\left(\dfrac{x}{2}+\dfrac{2}{x}\right)^2\)

\(\Rightarrow\sqrt{x^2+\dfrac{1}{x^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{2}{x}\right)\)

Tương tự:

\(\sqrt{y^2+\dfrac{1}{y^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{y}{2}+\dfrac{2}{y}\right)\) ; \(\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{z}{2}+\dfrac{2}{z}\right)\)

Cộng vế: \(P\ge\dfrac{2}{\sqrt{17}}\left(\dfrac{x}{2}+\dfrac{y}{2}+\dfrac{z}{2}+\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\right)\)

\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right)\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{36}{x+y+z}\right)\)

\(P\ge\dfrac{1}{\sqrt{17}}\left(x+y+z+\dfrac{9}{4\left(x+y+z\right)}+\dfrac{135}{4\left(x+y+z\right)}\right)\)

\(P\ge\dfrac{1}{\sqrt{17}}\left(2\sqrt{\dfrac{9\left(x+y+z\right)}{4\left(x+y+z\right)}}+\dfrac{135}{4.\dfrac{3}{2}}\right)=\dfrac{3}{2}\sqrt{17}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)

Ta có:

\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)

Áp dụng BĐT Cosi ta có:

\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)

Cmtt:

\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)

\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)

\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)

Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\dfrac{1}{\sqrt{x}+2\sqrt{y}}\le\dfrac{1}{9}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{y}}\right)\)

Tương tự cho 2 BĐT trên ta có:

\(\dfrac{1}{3}VP\le\dfrac{1}{9}\cdot3\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)=\dfrac{1}{3}VT\)

Xảy ra khi \(x=y=z\)