1.Cho a,b,c,da,b,c,d là các số nguyên thỏa mãn a3+b3=2(c3−d3)a3+b3=2(c3−d3) . Chứng minh rằng a+b+c+d chia hết cho 3

2.Cho ba số dương a,b,c thỏa mãn abc=1. Chứng minh rằng 1a3(b+c)+1b3(c+a)+1c3(a+b)≥32

\(4b.ac+\left(a+c\right)^2\le4b.\dfrac{1}{4}\left(a+c\right)^2+\left(a+c\right)^2=\left(a+c\right)^2\left(b+1\right)\)

\(\Rightarrow T\ge\dfrac{1}{\left(a+c\right)^2}+\dfrac{1}{\left(a+b\right)^2}\ge\dfrac{1}{2\left(a^2+c^2\right)}+\dfrac{1}{2\left(a^2+b^2\right)}\ge\dfrac{4}{2\left(2a^2+b^2+c^2\right)}\)

\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\\ =\left(ab^2-a\right)\left(c^2-1\right)+\left(a^2b-b\right)\left(c^2-1\right)+\left(a^2c-c\right)\left(b^2-1\right)\\ =ab^2c^2-ab^2-ac^2+a+a^2bc^2-a^2b-bc^2+b+a^2b^2c-a^2c-b^2c+c\\ =abc\left(ab+bc+ac\right)-\left(a^2b+ab^2+ac^2+bc^2+a^2c+b^2c\right)+\left(a+b+c\right)\\ =abc\left(ab+bc+ca\right)+\left(a+b+c\right)+3abc-\left[\left(a^2b+ab^2+abc\right)+\left(b^2c+bc^2+abc\right)+\left(a^2c+ac^2+abc\right)\right]\\ =abc\left(ab+bc+ca\right)+abc+3abc-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\right]\\ =4abc+abc\left(ab+bc+ca\right)-\left(a+b+c\right)\left(ab+bc+ca\right)\\ =4abc+abc\left(ab+bc+ca\right)-abc\left(ab+bc+ca\right)=4abc\)

Lời giải:

Ta có:
\(a(b^2-1)(c^2-1)+b(a^2-1)(c^2-1)+c(a^2-1)(b^2-1)\)

\(=a(b^2c^2-b^2-c^2+1)+b(a^2c^2-a^2-c^2+1)+c(a^2b^2-a^2-b^2+1)\)

\(=(ab^2c^2+ba^2c^2+ca^2b^2)+(a+b+c)-[a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)]\)

\(=abc(ab+bc+ac)+abc-[ab(a+b)+bc(b+c)+ca(c+a)]\)

\(=abc(ab+bc+ca)+4abc-[ab(a+b+c)+bc(b+c+a)+ca(c+a+b)]\)

\(=abc(ab+bc+ca)+4abc-(a+b+c)(ab+bc+ac)\)

\(=abc(ab+bc+ca)+4abc-abc(ab+bc+ac)=4abc\)

Ta có đpcm.