\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{x^2+y^2+2}{\left(xy\right)^2+x^2+y^2+1}=1-\dfrac{\left(xy\right)^2-1}{\left(xy\right)^2+x^2+y^2+1}\ge1-\dfrac{\left(xy\right)^2-1}{\left(xy\right)^2+2xy+1}\)

\(\Rightarrow\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge1-\dfrac{\left(xy+1\right)\left(xy-1\right)}{\left(xy+1\right)^2}=1-\dfrac{xy-1}{xy+1}=\dfrac{2}{1+xy}\) (đpcm)

b. Tương tự câu a:

\(\dfrac{1}{1+x^2}+\dfrac{1}{1+z^2}\ge\dfrac{2}{1+zx}\) ; \(\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{2}{1+yz}\)

Cộng vế với vế và rút gọn:

\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{z+zx}\) (1)

Mà \(\left\{{}\begin{matrix}z\ge1\Rightarrow1+xy\le1+xyz\\y\ge1\Rightarrow1+zx\le1+xyz\\x\ge1\Rightarrow1+yz\le1+xyz\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+zx}\ge\dfrac{1}{1+xyz}+\dfrac{1}{1+xyz}+\dfrac{1}{1+xyz}=\dfrac{3}{1+xyz}\) (2)

TỪ (1); (2) \(\Rightarrowđpcm\)

a) Ta có: \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)

\(\Leftrightarrow\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}+\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\ge0\)

\(\Leftrightarrow\dfrac{\left(1+xy\right)-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{\left(1+xy\right)-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(xy-x^2\right)\left(1+y^2\right)+\left(xy-y^2\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\dfrac{xy+xy^3-x^2-x^2y^2+xy+x^3y-y^2-x^2y^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{2xy+xy\left(x^2+y^2\right)-2x^2y^2-x^2-y^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{xy\left(x^2-2xy+y^2\right)-\left(x^2-2xy+y^2\right)}{\left(1+xy\right)\left(1+y^2\right)\left(1+x^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{xy\left(x-y\right)^2-\left(x-y\right)^2}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(xy-1\right)}{\left(1+xy\right)\left(1+x^2\right)\left(1+y^2\right)}\ge0\)(luôn đúng)

=> Đẳng thức ban đầu được chứng minh.

P/s: Cái đoạn sau bạn bổ sung thêm vào là vì x và y lớn hơn bằng 1 nên xy-1 sẽ lớn hơn hoặc bằng 0 nhé, mình lười quá ngại chèn:vv.

Còn câu b bạn đợi mình nháp xíu.

\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}\ge\dfrac{2}{1+xy}\)

⇔ \(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)

⇔ \(\left(\dfrac{1+xy-\left(1+x^2\right)}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-\left(1+y^2\right)}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)

⇔ \(\left(\dfrac{1+xy-1-x^2}{\left(1+x^2\right)\left(1+xy\right)}\right)+\left(\dfrac{1+xy-1-y^2}{\left(1+y^2\right)\left(1+xy\right)}\right)\ge0\)

⇔ \(\dfrac{-x\left(x-y\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{-y\left(y-x\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

⇔ \(\dfrac{-x\left(x-y\right)\left(1+y^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)\left(1+x^2\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

=> -x(x-y)(1+y²)+y(x-y)(1+x²) ≥ 0

⇔ (x-y)[-x(1+y²)+y(1+x²)]≥0

⇔ (x-y)(-x-xy²+y+x²y) ≥0

⇔ (x-y)[-(x-y)+(x²y-y²x)] ≥ 0

⇔ (x-y)[-(x-y)+xy(x-y) ]≥ 0

⇔ (x-y)(x-y)(xy-1)≥ 0

⇔ (x-y)² (xy-1) ≥0 (luôn đúng ∀ xy ≥ 1)

=> đpcm

Chứng minh bằng phép biến đổi tương đương:

1.

\(\Leftrightarrow4+x+y\ge4\sqrt{x+y}\)

\(\Leftrightarrow x+y-4\sqrt{x+y}+4\ge0\)

\(\Leftrightarrow\left(\sqrt{x+y}-2\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

2.

\(\Leftrightarrow\dfrac{y+z}{xyz}\ge\dfrac{4}{x^2+yz}\)

\(\Leftrightarrow\left(y+z\right)\left(x^2+yz\right)\ge4xyz\)

\(\Leftrightarrow x^2y+x^2z+y^2z+z^2y-4xyz\ge0\)

\(\Leftrightarrow y\left(x^2+z^2-2xz\right)+z\left(x^2+y^2-2xy\right)\ge0\)

\(\Leftrightarrow y\left(x-z\right)^2+z\left(x-y\right)^2\ge0\) (đúng)

Sai đề kìa.

Bạn tham khảo: Câu hỏi của Ngoc An Pham - Toán lớp 9 | Học trực tuyến

Ta có \(xy\left(x-y\right)^2+\left(xy-1\right)^2\ge0\Leftrightarrow\left(2+2x+2y+x^2+y^2\right)\left(1+xy\right)\ge\left(1+2x+x^2\right)\left(1+2y+y^2\right)\Leftrightarrow\left[\left(1+x\right)^2+\left(1+y\right)^2\right]\left(1+xy\right)\ge\left(1+x\right)^2\left(1+y\right)^2\Leftrightarrow\dfrac{\left(1+x\right)^2+\left(1+y\right)^2}{\left(1+x\right)^2\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\Leftrightarrow\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\)

Dấu bằng xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\xy-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy ta có \(\dfrac{1}{\left(1+x\right)^2}+\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{1}{1+xy}\)

Nhân 2 vào cả 2 vế:

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

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

\(=\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\)

\(VP=\dfrac{2}{1+xy}\)

Trở về bài toán của Ther

b) \(x,y\ge1\Rightarrow xy\ge1\)

BĐT đã cho tương đương với:

\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)

\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)

\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)

BĐT cuối luôn đúng nên ta có đpcm

Đẳng thức xảy ra khi x=y hoặc xy=1

Chứng minh 1/(1+x)^2 + 1/(1+y)^2 luôn >= 1/(1+xy)? | Yahoo Hỏi & Đáp

Nhấn vào link đó!

\(M=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)

\(=\dfrac{yz\sqrt{x-1}}{xyz}+\dfrac{xz\sqrt{y-2}}{xyz}+\dfrac{xy\sqrt{z-3}}{xyz}\)

\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

Áp dụng BĐT AM-GM ta có:

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

\(\sqrt{y-2}=\dfrac{\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{y}{2\sqrt{2}}\)\(\Rightarrow\dfrac{\sqrt{y-2}}{y}\le\dfrac{y}{2\sqrt{2}}\cdot\dfrac{1}{y}=\dfrac{1}{2\sqrt{2}}\)

\(\sqrt{z-3}=\dfrac{\sqrt{3\left(z-3\right)}}{\sqrt{3}}\le\dfrac{z}{2\sqrt{3}}\)\(\Rightarrow\dfrac{\sqrt{z-3}}{z}\le\dfrac{z}{2\sqrt{3}}\cdot\dfrac{1}{z}=\dfrac{1}{2\sqrt{3}}\)

Cộng theo vế 3 BĐT trên ta có:

\(M\le\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\) (ĐPCM)

z đâu ra???