Ta có: \(\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\left(c+\frac{1}{c}\right)\)

\(=\left(ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\right)\left(c+\frac{1}{c}\right)\)

\(=\left[ab+\frac{1}{16ab}+\frac{15}{16ab}+\left(\frac{a}{b}+\frac{b}{a}\right)\right]\left(c+\frac{1}{c}\right)\)

\(\ge\left[2\sqrt{ab.\frac{1}{16ab}}+\frac{15}{4\left(a+b\right)^2}+2\sqrt{\frac{a}{b}.\frac{b}{a}}\right]\left(2\sqrt{c.\frac{1}{c}}\right)\)

\(\ge\frac{25}{2}\left(Đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2};c=1\)

nó chưa cho c dương kìa.

tớ chỉ bày cách giải thôi

cm (a-1)(b-1)(c-1)>0

vì a.b.c=1 => (1.0)+1=1

từ đó sẽ suy ra là (a-1)(b-1)(c-1)>0

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 \(\Leftrightarrow a=2;b=1\)

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

Do a ; b > 0 , áp dụng BĐT Cô - si cho 2 số dương , ta có :

\(A=\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\ge2\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)\)

\(\Rightarrow2\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2\right]\ge\left(a+\frac{1}{a}+b+\frac{1}{b}\right)^2\)

\(\Rightarrow2A\ge\left(1+\frac{1}{a}+\frac{1}{b}\right)^2\)

Vì a ; b > 0 \(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

\(\Rightarrow2A\ge\left(1+\frac{4}{a+b}\right)^2=\left(1+4\right)^2=25\)

\(\Rightarrow A\ge\frac{25}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

chiều dài tấm vải chính bằng tổng số mét vải đã bán (vì ở đề bài nói rằng ngày 3 bán nốt 40m)

a)\(a^4+16\ge2a^3+8a\)

\(\Leftrightarrow a^4-2a^3-8a+16\ge0\)

\(\Leftrightarrow\left(a-2\right)^2\left(a^2+2a+4\right)\ge0\)

\(\Leftrightarrow\left(a-2\right)^2\left(\left(a+1\right)^2+3\right)\ge0\)*Luôn đúng*

\("="\Leftrightarrow a=2\)

b)Cô si: \(a+b\ge2\sqrt{ab}\)

\(\frac{1}{a}+\frac{1}{b}\ge2\sqrt{\frac{1}{ab}}\)

Nhân theo vế 2 BĐT trên ta đc ĐPCM

\("="\Leftrightarrow a=b\)