b: \(A=\dfrac{x^2+4+1}{\sqrt{x^2+4}}=\sqrt{x^2+4}+\dfrac{1}{\sqrt{x^2+4}}>=2\sqrt{\sqrt{x^2+4}\cdot\dfrac{1}{\sqrt{x^2+4}}}=2\)

a: =>ab+ad+bc+cd>=ab+cd+2căn abcd

=>ad+cb-2căn abcd>=0

=>(căn ad-căn cb)^2>=0(luôn đúng)

\(\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+c\right)\left(b+d\right)}\)

\(\Leftrightarrow ab+cd+2\sqrt{abcd}\le ab+bc+cd+da\)

\(\Leftrightarrow bc+da\ge2\sqrt{abcd}\)

\(\Leftrightarrow bc+da-2\sqrt{abcd}\ge0\)

\(\Leftrightarrow\left(\sqrt{bc}-\sqrt{da}\right)^2\ge0\) đúng \(\forall a,b,c,d>0\)

1) Áp dụng BĐT bun-hi-a-cốp-xki ta có:

\(\left(a+d\right)\left(b+c\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)

\(\Leftrightarrow\sqrt{\left(a+d\right)\left(b+c\right)}\ge\sqrt{ab}+\sqrt{cd}\)( vì a,b,c,d dương )

Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)

Ta có BĐT \(\Leftrightarrow\left(\sqrt{ab}+\sqrt{cd}\right)^2\le\left(a+d\right)\left(b+c\right)\Leftrightarrow ab+cd+2\sqrt{abcd}\le ab+ac+bd+dc\)

\(\Leftrightarrow ac+bd\ge2\sqrt{abcd}\) (luôn đúng theo AM-GM)

p/s: mà cái BĐT bn cần chứng minh đó chính là BĐT Bunyakovsky đấy ^.^

Ta có : \(\sqrt{\left(a+b\right)\left(c+d\right)}\ge\sqrt{ac}+\sqrt{bd}\)

\(\Leftrightarrow\left(a+b\right)\left(c+d\right)\ge\left(\sqrt{ac}+\sqrt{bd}\right)^2\)

\(\Leftrightarrow ac+ad+bc+bd\ge ac+2\sqrt{acbd}+bd\)

\(\Leftrightarrow ad-2\sqrt{adbc}+bc\ge0\)

\(\Leftrightarrow\left(\sqrt{ad}-\sqrt{bc}\right)^2\ge0\)( luôn đúng )

Dấu "=" xảy ra khi : \(ad=bc\)

Vậy ... 

Sử dụng bất đẳng thức Bunhiacopxki ta có :

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

\(\ge\left(\sqrt{ac}+\sqrt{bd}\right)^2\)

\(< =>\sqrt{\left(a+b\right)\left(c+d\right)}\ge\sqrt{ac}+\sqrt{bd}\left(đpcm\right)\)

okey?

Bài 1, t nghĩ VP căn phải kéo dài hết

Áp dụng bđt bu nhi a, ta có 

\(\left(\sqrt{ab}+\sqrt{cd}\right)^2\le\left(a+d\right)\left(b+c\right)\Rightarrow\sqrt{ab}+\sqrt{cd}\le\sqrt{\left(a+d\right)\left(b+c\right)}\left(ĐPCM\right)\)

Bài 2, Áp dụng bài 1, ta có 

\(\left(a\sqrt{3a\left(a+2b\right)}+b\sqrt{3b\left(b+2a\right)}\right)\le\left(a^2+b^2\right)\left[3a\left(a+2b\right)+3b\left(b+2a\right)\right]\)

\(\le2\left(3a^2+6ab+3b^2+6ab\right)=2\left[3\left(a^2+b^2\right)+12ab\right]\le2\left(6+12ab\right)\)

Áp dụng bđt cô si, ta có 

\(a^2+b^2\ge2ab\Rightarrow2\ge2ab\Rightarrow12\ge12ab\)

=>(...)^2<=36 => ...<=6 (ĐPcM)

dấu = xảy ra <=> a=b=1

^_^