Câu hỏi của Kelbin Noo

Question

Cho a, b, c > 0. Chứng minh: $\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

Nguyễn Huy Tú · Accepted Answer

Áp dụng bất đẳng thức AM - GM ta có:

$\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{2}{2\sqrt{ab}}+\dfrac{2}{2\sqrt{bc}}+\dfrac{2}{2\sqrt{ac}}$

$=\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\le\dfrac{1}{\sqrt{a^2}}+\dfrac{1}{\sqrt{b^2}}+\dfrac{1}{\sqrt{c^2}}$

$=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

Dấu " = " xảy ra khi $a=b=c$

Vậy...Câu trả lời: Áp dụng bất đẳng thức AM - GM ta có:

$\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\le\dfrac{2}{2\sqrt{ab}}+\dfrac{2}{2\sqrt{bc}}+\dfrac{2}{2\sqrt{ac}}$

$=\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{bc}}+\dfrac{1}{\sqrt{ca}}\le\dfrac{1}{\sqrt{a^2}}+\dfrac{1}{\sqrt{b^2}}+\dfrac{1}{\sqrt{c^2}}$

$=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

Dấu " = " xảy ra khi $a=b=c$

Vậy...

Lightning Farron · Answer

Áp dụng BĐT $\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$ ta có:

$\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$. Tương tự cho 2 BĐT còn lại có:

$\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}$

Cộng theo vế 3 BĐT trên ta có:

$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}$

$\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}$

Đẳng thức xảy ra khi $a=b=c$Câu trả lời: Áp dụng BĐT $\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$ ta có:

$\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$. Tương tự cho 2 BĐT còn lại có:

$\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c};\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+c}$

Cộng theo vế 3 BĐT trên ta có:

$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}$

$\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}$

Đẳng thức xảy ra khi $a=b=c$