Bài 2:
- Áp dụng bất đẳng thức Caushy, ta có:
\(\dfrac{1}{\sqrt[3]{\left(a+3b\right).1.1}}+\dfrac{1}{\sqrt[3]{\left(b+3c\right).1.1}}+\dfrac{1}{\sqrt[3]{\left(c+3a\right).1.1}}\)
\(\ge\dfrac{1}{\dfrac{a+3b+1+1}{3}}+\dfrac{1}{\dfrac{b+3c+1+1}{3}}+\dfrac{1}{\dfrac{c+3a+1+1}{3}}\)
\(=3\left(\dfrac{1}{a+3b+2}+\dfrac{1}{b+3c+2}+\dfrac{1}{c+3a+2}\right)\left(1\right)\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(3\left(\dfrac{1}{a+3b+2}+\dfrac{1}{b+3c+2}+\dfrac{1}{c+3a+2}\right)\)
\(\ge3.\dfrac{\left(1+1+1\right)^2}{a+3b+2+b+3c+2+c+3a+2}\)
\(=\dfrac{27}{4\left(a+b+c\right)+6}\)
\(=\dfrac{27}{4.\dfrac{3}{4}+6}=3\left(2\right)\)
- Từ (1), (2) ta suy ra đpcm.
- Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 1:
- Áp dụng bất đẳng thức Caushy ta có:
\(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}=3\sqrt[3]{1^2}=3\)
- Ta có:\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\)
\(=\dfrac{a^4}{abc+2ab}+\dfrac{b^4}{abc+2bc}+\dfrac{c^4}{abc+2ca}\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(\dfrac{a^4}{abc+2ab}+\dfrac{b^4}{abc+2bc}+\dfrac{c^4}{abc+2ca}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)+3abc}\)
hay \(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\)\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)+3abc}\left(1\right)\)
Mặt khác: \(ab+bc+ca\le a^2+b^2+c^2\left(2\right)\)
- Từ (1), (2):
\(\Rightarrow\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)+3abc}=\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)+3}\)- Đặt \(a^2+b^2+c^2=x\ge3\). Khi đó ta có:
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\ge\dfrac{x^2}{2x+3}\)
- Ta có: \(\dfrac{x^2}{2x+3}=\dfrac{x^2+\dfrac{3}{2}x-\dfrac{3}{2}x-\dfrac{9}{4}+\dfrac{9}{4}}{2x+3}\)
\(=\dfrac{\dfrac{1}{2}x\left(2x+3\right)-\dfrac{3}{4}\left(2x+3\right)+\dfrac{9}{4}}{2x+3}\)
\(=\dfrac{x}{2}-\dfrac{3}{4}+\dfrac{9}{4\left(2x+3\right)}\)
\(=\left[\dfrac{2x+3}{36}+\dfrac{9}{4\left(2x+3\right)}\right]+\dfrac{x}{2}-\dfrac{x}{18}-\dfrac{3}{4}-\dfrac{1}{12}\)
\(=\left[\dfrac{2x+3}{36}+\dfrac{9}{4\left(2x+3\right)}\right]+\dfrac{4}{9}x-\dfrac{5}{6}\left(3\right)\)
- Áp dụng bất đẳng thức Caushy ta có:
\(\left[\dfrac{2x+3}{36}+\dfrac{9}{4\left(2x+3\right)}\right]+\dfrac{4}{9}x-\dfrac{5}{6}\)
\(\ge2\sqrt{\dfrac{2x+3}{36}.\dfrac{9}{4\left(2x+3\right)}}+\dfrac{4}{9}.3-\dfrac{5}{6}\)
\(=2.\dfrac{1}{4}+\dfrac{4}{3}-\dfrac{5}{6}=1\left(4\right)\)
- Từ (3), (4) \(\Rightarrow\dfrac{x^2}{2x+3}\ge1\)
\(\Rightarrow\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b^3}{c\left(a+2\right)}+\dfrac{c^3}{a\left(b+2\right)}\ge\dfrac{x^2}{2x+3}\ge1\left(đpcm\right)\)
- Dấu "=" xảy ra khi \(a=b=c=1\)
\(1;\)
\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3}{3.9}}=a\Rightarrow\dfrac{a^3}{b\left(c+2\right)}\ge a-\dfrac{b}{3}-\dfrac{c+2}{9}\)
\(tương\) \(tự\Rightarrow\dfrac{b^3}{c\left(a+2\right)}\ge b-\dfrac{c}{3}-\dfrac{a+2}{3};\dfrac{c^3}{a\left(b+2\right)}\ge c-\dfrac{a}{3}-\dfrac{b+2}{3}\)
\(\Rightarrow\Sigma\dfrac{a^3}{b\left(c+2\right)}\ge a+b+c-\left(\dfrac{2\left(a+b+c\right)}{3}\right)=\dfrac{a+b+c}{3}\ge\dfrac{3\sqrt[3]{abc}}{3}=1\)
\(dấu:="\Leftrightarrow a=b=c=1\)
Bài 3:
- Áp dụng bất đẳng thức Caushy, ta có:
\(\dfrac{ab}{1-c^2}+\dfrac{bc}{1-a^2}+\dfrac{ca}{1-b^2}\)
\(\le\dfrac{\dfrac{\left(a+b\right)^2}{4}}{\left(1-c\right)\left(1+c\right)}+\dfrac{\dfrac{\left(b+c\right)^2}{4}}{\left(1-a\right)\left(1+a\right)}+\dfrac{\dfrac{\left(c+a\right)^2}{4}}{\left(1-b\right)\left(1+b\right)}\)
\(=\dfrac{1}{4}\left[\dfrac{\left(a+b\right)^2}{\left(a+b\right)\left(1+c\right)}+\dfrac{\left(b+c\right)^2}{\left(b+c\right)\left(1+a\right)}+\dfrac{\left(c+a\right)^2}{\left(c+a\right)\left(1+b\right)}\right]\)
\(=\dfrac{1}{4}\left(\dfrac{a+b}{1+c}+\dfrac{b+c}{1+a}+\dfrac{c+a}{1+b}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{1-c}{1+c}+\dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}\right)\)
\(=\dfrac{1}{4}\left[3-2.\left(\dfrac{c}{1+c}+\dfrac{a}{1+a}+\dfrac{b}{1+b}\right)\right]\left(1\right)\)
- Ta có: \(\dfrac{c}{1+c}+\dfrac{a}{1+a}+\dfrac{b}{1+b}\)
\(=1-\dfrac{1}{1+c}+1-\dfrac{1}{1+a}+1-\dfrac{1}{1+b}\)
\(=3-\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c+3}=\dfrac{9}{1+3}=\dfrac{9}{4}\)
- Khi đó ta có: \(\dfrac{c}{1+c}+\dfrac{a}{1+a}+\dfrac{b}{1+b}\le3-\dfrac{9}{4}=\dfrac{3}{4}\left(2\right)\)
- Từ (1), (2) suy ra:
\(\dfrac{ab}{1-c^2}+\dfrac{bc}{1-a^2}+\dfrac{ca}{1-b^2}\le\dfrac{1}{4}.\left(3-2.\dfrac{3}{4}\right)=\dfrac{3}{8}\left(đpcm\right)\)
- Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
