Ta có \(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

Khi đó \(abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge a+b+c\)

<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{a+b+c}{abc}\)

<=> \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}\)

<=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}\ge\frac{2}{bc}+\frac{2}{ac}+\frac{2}{ab}\)

<=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{bc}-\frac{2}{ac}-\frac{2}{ab}\ge0\)

<=> \(\left(\frac{1}{a^2}-\frac{2}{ac}+\frac{1}{c^2}\right)+\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)\ge0\)

<=> \(\left(\frac{1}{a}-\frac{1}{c}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2\ge0\left(\text{đúng }\forall a;b;c>0\right)\)

=> ĐPCM (Dấu "=" xảy ra <=> a = b = c) 

ta có :

\(\hept{\begin{cases}AB^2=BD.BC=9\left(9+16\right)=225\\AC^2=CD.CB=16\left(16+9\right)=400\end{cases}}\Leftrightarrow\hept{\begin{cases}AB=15\\AC=20\end{cases}}\)

nên diện tích ABC là : \(\frac{1}{2}AB.AC=\frac{1}{2}.15.20=150cm^2\)

Với x > = 0 ; \(x\ne1\)

\(A=\left(\frac{1}{1-\sqrt{x}}+\frac{1}{1+\sqrt{x}}\right)\left(\frac{1}{1-\sqrt{x}}-\frac{1}{1+\sqrt{x}}\right)+\frac{1}{1-\sqrt{x}}\)

\(=\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{1-x}\right)\left(\frac{1+\sqrt{x}-1+\sqrt{x}}{1-x}\right)+\frac{1}{1-\sqrt{x}}\)

\(=\frac{2\left(2\sqrt{x}\right)}{\left(1-x\right)^2}+\frac{1}{1-\sqrt{x}}=\frac{4\sqrt{x}}{\left(x-1\right)^2}-\frac{\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(x-1\right)^2}\)

\(=\frac{4\sqrt{x}-\left(\sqrt{x}+1\right)\left(x-1\right)}{\left(x-1\right)^2}\)

Ta có:

$a,b,c,d\in N$

Và $a.b.c.d=1$

$\iff a=b=c=d=1$

$\iff a^2+b^2+c^2+d^2+ab+cd=1+1+1+1+1+1=6$

$\Rightarrow đpcm$

Cho thêm a,b,c dương nữa nhé

Giải

Áp dụng BĐT AM-GM ta có:

\(a^2+b^2+c^2+d^2+ab+cd\)

\(>\)hoặc \(=\)\(6\)\(\sqrt{a^2b^2c^2d^2.ab.cd}\)

\(=6\sqrt[6]{\left(abcd\right)^3=6\sqrt[6]{1}=6\left(abcd=1\right)}\)

Đẳng thức xảy ra khi a=b=c=d=1

Theo BĐT Cauchy ta có : 

\(a^4+b^4+c^4+d^4\ge4\sqrt[4]{a^4b^4c^4d^4}=4abcd\)

Dấu ''='' xảy ra khi a = b = c = d = 1