\(a+b=1\Rightarrow b=1-a\)

\(a^3+b^3=a^3+\left(1-a\right)^3=3a^2-3a+1=3\left(a-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\) (đpcm)

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

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

\(x\ge2y\Rightarrow\frac{x}{y}\ge2\)

\(\Rightarrow M\ge2+\frac{1}{2}=\frac{5}{2}\)

GTNN của M là \(\frac{5}{2}\)khi \(a=2y\)

\(\frac{x}{y}>=2\)=>\(\frac{y}{x}=< \frac{1}{2}\)

\(M=\frac{x}{y}+\frac{y}{x}=\frac{x}{y}+\frac{4y}{x}-\frac{3y}{x}\)

ta có \(\frac{x}{y}+\frac{4y}{x}>=4\)(cô si)(1)

\(-\frac{3y}{x}>=-\frac{3}{2}\)(2)

cộng 1 với 2=>M>=5/2

xảy ra dâu = khi x/y=2

Đinh Khắc Duy : ngược dấu nha: x/y >= 2 suy ra \(\frac{1}{\frac{x}{y}}\le\frac{1}{2}\) nhá,và như thế ko c/m đc vì có 2 dấu

Ta có \(a\ge1;b\ge1\Rightarrow a\cdot b\ge1\) (1)

\(\Rightarrow\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)>0\) (2)

Từ (1);(2)\(\Rightarrow\dfrac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+ab\right)\left(1+a^2\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{b-a}{1+ab}\left(\dfrac{b^2\cdot a-a^2b-b+a}{\left(1+a^2\right)\left(1+b^2\right)}\right)\ge0\)

\(\Leftrightarrow\dfrac{b-a}{1+ab}\left(\dfrac{a}{1+a^2}-\dfrac{b}{1+b^2}\right)\ge0\)

\(\Leftrightarrow\dfrac{ab-a^2}{\left(1+ab\right)\left(1+a^2\right)}-\dfrac{b^2-ab}{\left(1+ab\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{ab-a^2+1-1}{\left(1+ab\right)\left(1+a^2\right)}-\dfrac{b^2-1-ab+1}{\left(1+ab\right)\left(1+b^2\right)}\ge0\)

\(\Leftrightarrow\dfrac{1}{1+a^2}-\dfrac{1}{1+ab}+\dfrac{1}{1+b^2}-\dfrac{1}{1+ab}\ge0\)

\(\Rightarrow\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}\ge\dfrac{2}{1+ab}\) (đpcm)

P/S: x thay = a , y thay = b nha

Hiển nhiên \(c\left(c+1\right)>a\left(a+1\right)\Rightarrow c>a\ge b\)

Nếu \(c\ge2a\Rightarrow c\left(c+1\right)\ge2a\left(2a+1\right)=4a^2+2a\)

Mà \(a\left(a+1\right)+b\left(b-1\right)\le a\left(a+1\right)+a\left(a-1\right)=2a^2\)

\(\Rightarrow2a^2\ge4a^2+2a\Rightarrow2a^2+2a\le0\) (vô lý)

\(\Rightarrow c< 2a\)

Ta có:

\(4a\left(a+1\right)+4b\left(b-1\right)+1=4c\left(c+1\right)+1\)

\(\Leftrightarrow4a\left(a+1\right)+\left(2b-1\right)^2=\left(2c+1\right)^2\)

\(\Leftrightarrow4a\left(a+1\right)=\left(2c+1\right)^2-\left(2b-1\right)^2\)

\(\Leftrightarrow a\left(a+1\right)=\left(c-b+1\right)\left(c+b\right)\) (*)

Nếu \(c-b+1\ge a\Rightarrow\left(c-b+1\right)\left(c+b\right)>a\left(a+b\right)>a\left(a+1\right)\) (ktm)

\(\Rightarrow c-b+1< a\) \(\Rightarrow c-b+1\) ko có ước nguyên tố nào là a

\(\Rightarrow c+b⋮a\Rightarrow\dfrac{c+b}{a}\in Z\) (1)

Theo chứng minh ban đầu, ta có \(b\le a< c< 2a\)

\(\Rightarrow a< c+b< 2a+a=3a\Rightarrow1< \dfrac{c+b}{a}< 3\) (2)

(1);(2) \(\Rightarrow\dfrac{c+b}{a}=2\Rightarrow c+b=2a\)

Thế vào (*) \(\Rightarrow a+1=2\left(c-b+1\right)\Rightarrow2c-2b+1=a\)

\(\Rightarrow2\left(2a-b\right)-2b+1=a\Rightarrow3a-4b+1=0\)

\(\Rightarrow3\left(a-1\right)=4\left(b-1\right)\)

\(\Rightarrow b-1⋮3\Rightarrow b-1=3k\Rightarrow b=3k+1\)

\(\Rightarrow a=4k+1\)

\(\Rightarrow c=2a-b=5k+1\)

\(\Rightarrow A=3\left(5k+1\right)-5\left(3k+1\right)=-2\)

\(\Leftrightarrow\left(2+a^2+b^2\right)\left(1+ab\right)\ge2\left(1+a^2\right)\left(1+b^2\right)\)

\(\Leftrightarrow2+2ab+a^2+b^2+ab\left(a^2+b^2\right)\ge2+2a^2+2b^2+2a^2b^2\)

\(\Leftrightarrow ab\left(a^2+b^2-2ab\right)-\left(a^2+b^2-2ab\right)\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\) (luôn đúng với mọi \(a\ge1;b\ge1\))

Cách khác:

\(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)+\left(\frac{1}{1+b^2}-\frac{1}{1+ab}\right)\ge0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}+\frac{b\left(a-b\right)}{\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)\left[b\left(1+a^2\right)-a\left(1+b^2\right)\right]}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\) (luôn đúng).

\(\dfrac{9}{4}=ab+a+b+1\le\dfrac{1}{4}\left(a+b\right)^2+a+b+1\)

\(\Leftrightarrow\left(a+b\right)^2+4\left(a+b\right)-5\ge0\)

\(\Leftrightarrow\left(a+b-1\right)\left(a+b+5\right)\ge0\)

\(\Leftrightarrow a+b-1\ge0\) (do \(a+b+5>0\))

\(\Rightarrow a+b\ge1\)

b.

\(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\ge\dfrac{1}{2}.1^2=\dfrac{1}{2}\) (đpcm)