Cho a,b,c>0 tm a+b+c=3
Tìm min `P=(b\sqrt{b})/(\sqrt{2a+b+c})+(c\sqrt{c})/(\sqrt{2b+c+a})+(a\sqrt{a})/(\sqrt{2c+a+b)}`
Dùng BĐT Cauchy 3 số để làm
Cho a,b,c>0 tm a+b+c=3
Tìm min `P=(b\sqrt{b})/(\sqrt{2a+b+c})+(c\sqrt{c})/(\sqrt{2b+c+a})+(a\sqrt{a})/(\sqrt{2c+a+b)}`
Dùng BĐT Cauchy 3 số để làm
$\dfrac{b.\sqrt[]b}{\sqrt[]{2a+b+c}}$
$=\dfrac{b}{\sqrt[]{\dfrac{2a+b+c}{b}}}$
$=\dfrac{b}{\sqrt[]{\dfrac{a+3}{b}}}$
Áp dụng bất đẳng thức Cauchy có:
$\dfrac{1}{2}.\sqrt[]{4.\dfrac{a+3}{b}}≤\dfrac{1}{2}.\dfrac{4+\dfrac{a+3}{b}}{2}=\dfrac{a+4b+3}{4b}$
$⇒\dfrac{b}{\sqrt[]{\dfrac{a+3}{b}}}≥\dfrac{4b^2}{a+4b+3}$
Hay $\dfrac{b.\sqrt[]b}{\sqrt[]{2a+b+c}}≥\dfrac{4b^2}{a+4b+3}$
Tương tự: $\dfrac{c.\sqrt[]c}{\sqrt[]{2b+c+a}}≥\dfrac{4c^2}{b+4c+3}$
$\dfrac{a.\sqrt[]a}{\sqrt[]{2c+a+b}}≥\dfrac{4a^2}{c+4a+3}$
Nên $P≥\dfrac{4b^2}{a+4b+3}+\dfrac{4c^2}{b+4c+3}+\dfrac{4a^2}{c+4a+3}$
Áp dụng bất đẳng thức Svacxo có:
$\dfrac{4b^2}{a+4b+3}+\dfrac{4c^2}{b+4c+3}+\dfrac{4a^2}{c+4a+3}≥4.\dfrac{(a+b+c)^2}{5.(a+b+c)+9}=4.\dfrac{3^2}{5.3+9}=\dfrac{3}{2}$
Hay $P≥\dfrac{3}{2}$
Dấu $=$ xảy ra $⇔a=b=c=1$
Đáp án + giải thích các bước giải:
$\dfrac{b\sqrt{b}}{\sqrt{2a+b+c}}=\dfrac{\sqrt{b^3}}{\sqrt{a+3}}=\dfrac{\sqrt{b^3}}{\sqrt{a+3}}+\dfrac{a+3}{16}-\dfrac{a+3}{16}=\dfrac{1}{2} (\dfrac{\sqrt{b^3}}{\sqrt{a+3}}+\dfrac{\sqrt{b^3}}{\sqrt{a+3}}+\dfrac{a+3}{8})-\dfrac{a+3}{16}$
Áp dụng bất đẳng thức Cô-si cho ba số:
$\dfrac{1}{2} (\dfrac{\sqrt{b^3}}{\sqrt{a+3}}+\dfrac{\sqrt{b^3}}{\sqrt{a+3}}+\dfrac{a+3}{8})-\dfrac{a+3}{16} \ge \dfrac{1}{2} 3\sqrt[3]{\dfrac{\sqrt{b^3}}{\sqrt{a+3}} . \dfrac{\sqrt{b^3}}{\sqrt{a+3}} . \dfrac{a+3}{8}}-\dfrac{a+3}{16} =\dfrac{1}{2} . 3 . \dfrac{b}{2}-\dfrac{a+3}{16}=\dfrac{3}{4}b-\dfrac{a+3}{16}$
Tương tự có:
`P>=3/4 (a+b+c) – (a+3+b+3+c+3)/16=3/4. 3 -12/16=3/2`
Dấu bằng xảy ra khi `a=b=c=1`