Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)\(+2\left(ab^2c+abc^2+a^2bc\right)\)
=\(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\)
=\(a^2b^2+b^{2^2}c^2+c^2a^2+2abc.0\)
=\(a^2b^2+b^2c^2+c^2a^2\)
b) \(a+b+c=0\)=>\(\left(a+b+c\right)^2=0\)
<=>\(a^2+b^2+c^2+2\left(ab+bc+xa\right)=0\)
<=>\(a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)
<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)\(=4\left(ab+bc+ca\right)^2\)
Do \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)
=>\(a^4+b^4+c^4+2\left(ab+bc+ca\right)^2\)\(=4\left(ab+bc+ca\right)^2\)
=>\(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(=a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow2\left(ab+bc+ac\right)=3\left(a^2+b^2+c^2\right)-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow2\left(ab+bc+ac\right)=2\left(a^2+b^2+c^2\right)\)
\(\Rightarrow ab+bc+ac=a^2+b^2+c^2\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
Lần lượt áp dụng bất đẳng thức Cô - si có 3 và 4 số, ta có:
\(\frac{a}{18}+\frac{b}{24}+\frac{2}{ab}\ge3.\sqrt[3]{\frac{a}{18}.\frac{b}{24}.\frac{2}{ab}}=\frac{1}{2}\)
\(\frac{a}{9}+\frac{c}{6}+\frac{2}{ac}\ge3.\sqrt[3]{\frac{a}{9}.\frac{c}{6}.\frac{2}{ac}}=1\)
\(\frac{b}{16}+\frac{c}{8}+\frac{2}{bc}\ge3.\sqrt[3]{\frac{b}{16}.\frac{c}{8}.\frac{2}{bc}}=\frac{3}{4}\)
\(\frac{a}{9}+\frac{b}{12}+\frac{c}{6}+\frac{8}{abc}\ge4.\sqrt[4]{\frac{a}{9}.\frac{b}{12}.\frac{c}{6}.\frac{8}{abc}}=\frac{4}{3}\)
\(\frac{13a}{18}+\frac{13b}{24}\ge2\sqrt{\frac{13a}{18}.\frac{13b}{24}}\ge2\sqrt{\frac{13.13.12}{18.24}}=\frac{13}{3}\)
\(\frac{13c}{24}+\frac{13b}{48}\ge2\sqrt{\frac{13c}{24}.\frac{13b}{48}}\ge2\sqrt{\frac{13.13.8}{24.48}}=\frac{13}{6}\)
Cộng vế với vế ta có:
\(a+b+c+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{8}{abc}\ge\frac{121}{12}\)