3: =>a^2c^2+a^2d^2+b^2c^2+b^2d^2>=a^2c^2+2abcd+b^2d^2

=>a^2d^2-2abcd+b^2c^2>=0

=>(ad-bc)^2>=0(luôn đúng)

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\((a^2+1)(c^2+4)=(a^2+1^2)(c^2+2^2)\geq (ac+2)^2\)

\((b^2+2)(d^2+8)=(b^2+\sqrt{2}^2)(d^2+\sqrt{8}^2)\geq (bd+\sqrt{2.8})^2=(bd+4)^2\)

Nhân theo vế 2 BĐT trên thu được đpcm

Dấu "=" xảy ra khi: \(\left\{\begin{matrix} \frac{a}{c}=\frac{1}{2}\\ \frac{b}{d}=\frac{\sqrt{2}}{\sqrt{8}}=\frac{1}{2}\end{matrix}\right.\)

Áp

Áp dụng BĐT Bunhiacopski với 2 bộ số( a,b,c,d)(1,1,1,1) ta có:

(a.1+b.1+c.1+d.1)²\(\le\)(1+1+1+1)(\(a^2+b^2+c^2+d^2)\)

<=>\(4(a^2+b^2+c^2+d^2)\ge2^2 \)

<=>\(a^2+b^2+c^2+d^2\ge1\)

Dấu bằn xảy ra<=> a=b=c=d=\(\frac{1}{2} \)

Biến đổi tương đương:

\(\Leftrightarrow a^6+a^5b+ab^5+b^6>a^6+a^4b^2+a^2b^4+b^6\)

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

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

\(\Leftrightarrow ab\left(a-b\right)\left(a^3-b^3\right)\ge0\)

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

\(\dfrac{\tan A}{\tan B}=\dfrac{\sin A}{\cos A}.\dfrac{\cos B}{\sin B}=\dfrac{\dfrac{a.\sin B}{b}\left(\dfrac{a^2+c^2-b^2}{2ac}\right)}{\dfrac{b^2+c^2-a^2}{2bc}.\sin B}=\dfrac{\dfrac{\sin B.\left(a^2+c^2-b^2\right)}{2bc}}{\dfrac{\sin B.\left(b^2+c^2-a^2\right)}{2bc}}=\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}\)

Câu 1: 

\(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

\(=\dfrac{\left(a+b+c\right)\cdot\left(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\right)}{2}\)

\(=\dfrac{\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]}{2}>=0\)

=>\(a^3+b^3+c^3>=3abc\)