`\sqrta+\sqrtb >= \sqrt(a+b)`

`<=> a+b+2\sqrt(ab) >= a+b`

`<=> 2\sqrt(ab) >= 0`

`<=> ab>=0` (Đúng)

(`a>=0 ; b>= 0 => ab >=0 forall a,b`)

`=>` ĐPCM.

Ta có:

\(\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\left(\sqrt{a}\right)^2+\sqrt{a}.\sqrt{b}+\sqrt{b}.\sqrt{a}+\left(\sqrt{b}\right)^2\)

\(=a+b+2\sqrt{a}.\sqrt{b}\)

\(=\left(\sqrt{a+b}\right)^2+2\sqrt{a}.\sqrt{b}\)

Vì \(\sqrt{a}\ge0,\sqrt{b}\ge0\) nên \(2\sqrt{a}.\sqrt{b}\ge0\) cho nên

\(\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a+b}\right)^2=2\sqrt{a}.\sqrt{b}\ge0\).

Tức là \(\left(\sqrt{a}+\sqrt{b}\right)^2\ge\left(\sqrt{a+b}\right)^2,\) suy ra \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)

Đẳng thức \(\sqrt{a}+\sqrt{b}=\sqrt{a+b}\) xảy ra chỉ khi \(\sqrt{a}.\sqrt{b}=0\)

tức là khi \(\sqrt{a}=0\) hoặc \(\sqrt{b}=0\), hay là \(a=0\) hoặc \(b=0\).

bài này hay đấy

Áp dụng BĐT Cô-si cho 3 số không âm, ta có :

\(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\ge3\sqrt[3]{\frac{1+\sqrt{a}}{1+\sqrt{b}}.\frac{1+\sqrt{b}}{1+\sqrt{c}}.\frac{1+\sqrt{c}}{1+\sqrt{a}}}=3\)

Chứng minh \(\frac{1+\sqrt{a}}{1+\sqrt{b}}+\frac{1+\sqrt{b}}{1+\sqrt{c}}+\frac{1+\sqrt{c}}{1+\sqrt{a}}\le3+a+b+c\)( 1 )

đặt \(\sqrt{a}=x;\sqrt{b}=y;\sqrt{c}=z\)( x,y,z \(\ge\)0 )

do a,b,c là số nguyên 

Nếu a = b = c = 0 thì x = y = z = 0 nên ( 1 ) đúng

Nếu a,b,c không đồng thời bằng 0 \(\Rightarrow\)x+ y + z \(\ge\)1

Ta có : VT ( 1 ) 

\(\Leftrightarrow\frac{\left(1+x\right)\left(1+y\right)-\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)\left(1+z\right)-\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)\left(1+x\right)-\left(1+z\right)x}{1+z}\)

\(=3+x+y+z-\left[\frac{\left(1+x\right)y}{1+y}+\frac{\left(1+y\right)z}{1+z}+\frac{\left(1+z\right)x}{1+x}\right]\)

\(\le3+x+y+z-\frac{\left(1+x\right)y+\left(1+y\right)z+\left(1+z\right)x}{1+x+y+z}=3+x+y+z-\frac{x+y+z+xy+yz+xz}{1+x+y+z}\)

\(=3+\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le3+x^2+y^2+z^2\)

Cần chứng minh : \(\frac{x^2+y^2+z^2+xy+yz+xz}{1+x+y+z}\le x^2+y^2+z^2\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)

Mà \(\left(x+y+z\right)\left(x^2+y^2+z^2\right)\ge1.\left(x^2+y^2+z^2\right)\ge xy+yz+xz\)

suy ra đpcm

a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)