\(\sqrt{3\left(a^2+6\right)}\ge\sqrt{2}\left(a+b\right)\)

\(\Leftrightarrow\sqrt{3\left(2a^2+b^2\right)}\ge\sqrt{2}\left(a+b\right)\)

\(\Leftrightarrow6a^2+3b^2\ge2a^2+4ab+2b^2\)

\(\Leftrightarrow4a^2-4ab+b^2\ge0\)

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

=> ĐPCM

Lời giải:

Từ ĐKĐB kết hợp BĐT Bunhiacopxky:
\(3(a^2+6)=3(a^2+a^2+b^2)=(1+2)(2a^2+b^2)\geq (\sqrt{2}a+\sqrt{2}b)^2\)

\(\Rightarrow \sqrt{3(a^2+6)}\geq \sqrt{2}(a+b)\) (đpcm)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} a,b>0\\ a^2+b^2=6\\ \frac{1}{\sqrt{2}a}=\frac{\sqrt{2}}{b}\end{matrix}\right.\) hay $a=\sqrt{\frac{6}{5}}; b=2\sqrt{\frac{6}{5}}$

ta có: \(\sqrt{3\left(a^2+b^2\right)}\ge\left(a+b\right)^2\sqrt{2}\)

\(\Leftrightarrow3\left(2a^2+b^2\right)\ge2\left(a+b\right)^2\)(vì a²+b²=6)

\(\Leftrightarrow6a^2+3b^2\ge2a^2+4ab+2b^2\)

\(\Leftrightarrow4a^2-4ab+b^2\ge0\\ \Leftrightarrow\left(2a-b\right)^2\ge0\)

(luôn đúng với mọi a;bdương)

=> đpcm

Bình phương 2 vế ta được 

3a² + 18 - 2a² - 4ab - 2b² \(\ge\)0

<=> a² - 2b² - 4ab + 3( a² + b²) \(\ge0\)

<=> 4a² - 4ab + b² \(\ge0\)

<=> (2a - b)² \(\ge0\)(đúng)

toán lớp 1 ??? giỡn quài , phi logic :3

Ap dung bdt AM-GM cho 2 so ko am A,B ta co 

\(\sqrt{A}+\sqrt{B}\)\(\le\)\(2\sqrt{\frac{A+B}{2}}\)

VP =\(\sqrt{AB}.\left(\sqrt{A}+\sqrt{B}\right)\le\frac{A+B}{2}.2\sqrt{\frac{A+B}{2}}\)

    =>VP² \(\le4.\frac{\left(A+B\right)^3}{4}=\left(A+B\right)^3\left(3\right)\)

Tu (2),(3) => DPCM

Dấu "=" xảy ra khi a=b=0 hoặc a=b=4, vẫn xét dấu "=" được nhé Đinh Nhật Minh 

Áp dụng BĐT Bunhiacopxki:

\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ac+bc\right)^2}=ac+bc\)

CMTT : \(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ad+bd\)

Ta có :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)

Áp dụng BĐT Bunhiacopxki:

(�2+�2)(�2+�2)≥(��+��)2=��+��(a2+c2)(b2+c2)​≥(ac+bc)2​=ac+bc

CMTT : (�2+�2)(�2+�2)≥��+��(a2+d2)(b2+d2)​≥ad+bd

Ta có :(�2+�2)(�2+�2)+(�2+�2)(�2+�2)≥��+��+��+��=(�+�)(�+�)(a2+c2)(b2+c2)​+(a2+d2)(b2+d2)​≥ac+bc+ad+bd=(a+b)(c+d)

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

\(VT=\sqrt{\left[a^2-2a+1+a^2+2a+1\right]\left[b^2+2bc+c^2+b^2c^2-2bc+1\right]}\)

\(VT=\sqrt{\left[\left(1-a\right)^2+\left(a+1\right)^2\right]\left[\left(bc-1\right)^2+\left(b+c\right)^2\right]}\)

Bunhiacopxki:

\(VT\ge\left(1-a\right)\left(bc-1\right)+\left(a+1\right)\left(b+c\right)=\left(1+a\right)\left(1+b\right)\left(1+c\right)-2\left(1+abc\right)\)