Áp dụng BĐT Bunhiacopxki với 2 dãy số: a;b;c và 1;1;1. Ta có:
\(\left(a^2+b^2+c^2\right)\left(1+1+1\right)\ge\left(a\times1+b\times1+c\times1\right)^2\)
\(\left(a^2+b^2+c^2\right)\times3\ge\left(a+b+c\right)^2\)
\(\left(a^2+b^2+c^2\right)\times3\ge9\)
\(\left(a^2+b^2+c^2\right)\ge3\)
Dấu bằng xảy ra <=> \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}<=>a=b=c\)
Áp dụng bất đẳng thức cô-si, ta có:
\(a^2+b^2\ge2ab\)
\(b^2+c^2\ge2bc\)
\(c^2+a^2\ge2ca\)
=>\(a^2+b^2+b^2+c^2+c^2+a^2\ge2ab+2bc+2ca\)
=>\(2.\left(a^2+b^2+c^2\right)\ge ab+ab+bc+bc+ca+ca\)
=>\(2.\left(a^2+b^2+c^2\right)+a^2+b^2+c^2\ge ab+ab+bc+bc+ca+ca+a^2+b^2+c^2\)
=>\(3.\left(a^2+b^2+c^2\right)\ge\left(a^2+ab+ca\right)+\left(ab+b^2+bc\right)+\left(ca+bc+c^2\right)\)
=>\(3.\left(a^2+b^2+c^2\right)\ge a.\left(a+b+c\right)+b.\left(a+b+c\right)+c.\left(a+b+c\right)\)
=>\(3.\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right).\left(a+b+c\right)\)
=>\(3.\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Vì a+b+c=3
=>\(3.\left(a^2+b^2+c^2\right)\ge3^2\)
=>\(3.\left(a^2+b^2+c^2\right)\ge9\)
=>\(a^2+b^2+c^2\ge3\)
Dấu "=" xảy ra khi: a=b=c=1
=>ĐPCM
Dùng Bunhiacopxki nhanh hơn nhiều mà bạn
