Đặt \(\hept{\begin{cases}b-1=x\\a-1=y\end{cases}\left(x,y>0\right)}\)

Ta có : \(P=\frac{\left(x+1\right)^2}{y}+\frac{\left(y+1\right)^2}{x}\)

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

\(=\left(\frac{x^2}{y}+\frac{y^2}{x}\right)+2.\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{x}+\frac{1}{y}\)

\(\ge\frac{\left(x+y\right)^2}{x+y}+2\cdot2.\sqrt{\frac{x}{y}\cdot\frac{y}{x}}+\frac{4}{x+y}\)

\(=\left(x+y+\frac{4}{x+y}\right)+4\)

\(\ge2\sqrt{\left(x+y\right)\cdot\frac{4}{x+y}}+4=2.2+4=8\)

Do đó \(P\ge8\)

Dấu "=" xảy ra khi \(a=b=2\)

Thên 1 cách: \(P=\frac{a^2}{b-1}+\frac{b^2}{a-1}\ge\frac{\left(a+b\right)^2}{a+b-2}\)

Đặt a + b =t > 2 

=> \(P=\frac{t^2}{t-2}=\frac{t^2-4+4}{t-2}=\left(t+2\right)+\frac{4}{t-2}=\left(t-2\right)+\frac{4}{t-2}+4\ge4+4=8\)

Dấu "=" xảy ra <=>a = b và  t - 2 = 2 <=> a = b = 2

Xét hiệu : \(\frac{a^3+b^3}{2}-\left(\frac{a+b}{2}\right)^3=\frac{4.\left(a^3+b^3\right)-\left(a+b\right)^3}{8}\)

\(=\frac{3.\left(a^3+b^3-a^2b-ab^2\right)}{8}\)

\(=\frac{3\left(a+b\right)\left(a-b\right)^2}{8}\ge0\) ( luôn đúng )

Do đó : \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)

Dấu "= xảy ra khi a=b.

1) \(VT=\frac{\sqrt{a}+\sqrt{b}}{2\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}-\sqrt{b}}{2\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2b}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}=VP\)(ĐPCM)

2) \(VT=\text{[}\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a+b-\sqrt{ab}\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\text{]}.\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)

\(=\frac{\left(a+b-\sqrt{ab}-\sqrt{ab}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}=\frac{\left(a-b\right)^2}{\left(a-b\right)^2}=1=VP\)(ĐPCM)

4) \(VT=\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a=VP\)(ĐPCM)

Trước hết ta sẽ chứng minh bổ đề phụ sau, với mọi a,b dương ta có: 

\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

Thật vậy  biến đổi tương đương ta đưa về \(\left(a-b\right)^2\left(a^2+ab+b^2\right)=0\)

BĐT này luôn đúng, thế thì

\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)

\(\Rightarrow\left(a^4+b^4\right)\ge\frac{\left(a+b\right)\left(a^3+b^3\right)}{2}\)

\(\frac{a^4+b^4}{a^3+b^3}\ge\frac{a+b}{2}\)

Như vậy ta có:

\(\hept{\begin{cases}\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\\\frac{y^4+z^4}{y^3+z^3}\ge\frac{y+z}{2}\\\frac{z^4+x^4}{z^3+x^3}\ge\frac{z+x}{2}\end{cases}}\)

\(\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^3+x^3}\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=1\)

Dấu '=' xảy ra khi x=y=z=1/3

Đặng Ngọc Quỳnh  không cần a,b rồi suy ra x,y, quá lòng vòng

Bạn tham khảo cách làm tại đây

 Câu hỏi của Pham Quoc Cuong - Toán lớp 8 - Học toán với OnlineMath