Question

[Nguyễn Thị Mừng]

Cho a,b,c là ba số dương thỏa mãn a+b+c=9.Chưng minh 1/a+1/b+1/c>1

Answer 1

[Xyz OLM]

Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge1\)(1)

<=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)(Vì a + b + c = 9)

<=> \(1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\ge9\) 

<=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)

Lại có \(\frac{a}{b}+\frac{b}{a}\ge2\)

<=>  \(\frac{a^2+b^2}{ab}\ge2\Leftrightarrow a^2+b^2\ge2ab\Leftrightarrow\left(a-b\right)^2\ge0\left(\text{đúng}\right)\)

Tương tự \(\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\\\frac{b}{c}+\frac{c}{b}\ge2\end{cases}}\)

<=> \(\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge6\)(đúng) 

=> (1) được chứng minh

Answer 2

[Nguyễn Hải Minh]

Áp dụng bđt Svac-xơ ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}=\frac{9}{9}=1\) ( Vì a+b+c=1)