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Ta có: \(P=1+\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)+\left(\frac{1}{a^3b^3}+\frac{1}{b^3c^3}+\frac{1}{a^3c^3}+\frac{1}{a^3b^3c^3}\right)\)
\(P\ge a+\frac{3}{abc}+\frac{3}{a^2b^2c^2}+\frac{1}{a^3b^3c^3}=\left(1+\frac{1}{abc}\right)^3\) (BĐT Cosi cho 3 số dương)
Theo BĐT Cosi \(abc\le\left(\frac{a+b+c}{3}\right)^3=8̸\)\(\Rightarrow abc\le8\Rightarrow\frac{1}{abc}\ge\frac{1}{8}\)
Vậy \(P\ge\left(1+\frac{1}{8}\right)^3=\frac{729}{512}\)
Dấu "=" xảy ra khi a=b=c=2
1. Áp dụng BĐT Cauchy dạng Engle, ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)
\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)
\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)
Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)
\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)
Áp dụng BĐT Cauchy cho a ; b dương
Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)
\(P=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ac}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\left(abc=1\right)\)
\(=\frac{1}{a^2\left(\frac{1}{c}+\frac{1}{b}\right)}+\frac{1}{b^2\left(\frac{1}{c}+\frac{1}{a}\right)}+\frac{1}{c^2\left(\frac{1}{b}+\frac{1}{a}\right)}\)
\(=\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{a}}\)
Đặt \(\left\{\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) suy ra \(xyz=1\). Khi đó:
\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Áp dụng BĐT AM-GM ta có:
\(\left\{\begin{matrix}\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\\\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\\\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\end{matrix}\right.\).Cộng theo vế ta có:
\(P+\frac{x+y+z}{2}\ge x+y+z\)
\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{3}{2}\left(x+y+z\ge3\sqrt[3]{xyz}=3\right)\)
Chứng minh bất đẳng thức \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Có: \(\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\left(\sqrt{x}^2+\sqrt{y}^2+\sqrt{z}^2\right)\ge\left(a+b+c\right)^2\) (Bunyakovsky)
\(\Leftrightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
abc = 1 => a^2.b^2.c^2 = 1
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}=\frac{a^2b^2c^2}{a^3\left(b+c\right)}+\frac{a^2b^2c^2}{b^3\left(c+a\right)}+\frac{a^2b^2c^2}{c^3\left(a+b\right)}\)
\(=\frac{\left(bc\right)^2}{ab+ac}+\frac{\left(ac\right)^2}{bc+ba}+\frac{\left(ab\right)^2}{ca+cb}\ge\frac{\left(ab+ac+bc\right)^2}{2\left(ab+ac+bc\right)}=\frac{\left(ab+ac+bc\right)}{2}\)
\(\ge\frac{3\sqrt[3]{ab.ac.bc}}{2}\)(Cauchy) \(=\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}a=b=c\\\frac{bc}{ab+ac}=\frac{ac}{bc+ba}+\frac{ab}{ca+cb}\Leftrightarrow\end{cases}a=b=c}\)
Mà abc=1 <=> a^3 = 1 <=> a=1 => b=c=a=1
https://diendantoanhoc.net/topic/80159-ch%E1%BB%A9ng-minh-frac1a2b3cfrac12a3bcfrac13bb2c-leqslant-frac316/
bạn tham khảo ở đây nhé
ta có A=\(\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}+\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}=\frac{a^2+b^2+c^2}{abc}+\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}\)
mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrow\frac{a^2+b^2+c^2}{abc}\ge\frac{ab+bc+ca}{abc}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow A\ge\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a^2}{2}+\frac{1}{2a}+\frac{1}{2a}+...\)
Áp dụng bđt co si ta có , \(\frac{a^2}{2}+\frac{1}{2a}+\frac{1}{2a}\ge\frac{1}{\sqrt{2}}\)
tương tự mấy cái kia rồi + vào thì A>=...
a+b+c=3
Mà: \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le1\Rightarrow abc\le1\Rightarrow\frac{1}{abc}\ge1\)(cô-si)
\(P=1+\frac{3}{a}+\frac{3}{b}+\frac{3}{c}+\frac{9}{ab}+\frac{9}{ac}+\frac{9}{bc}+\frac{27}{abc}\)
\(=1+3.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+9.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+\frac{27}{abc}\)
\(\ge1+9\sqrt[3]{\frac{1}{abc}}+27\sqrt[3]{\frac{1}{\left(abc\right)^2}}+\frac{27}{abc}\ge1+9\sqrt[3]{\frac{1}{1}}+27\sqrt[3]{\frac{1}{1}}+\frac{27}{1}=64\)
Vậy GTNN của P là 64 tại a=b=c=1