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áp dụng bdt côsi \(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{3}{b}\)
tuông tu \(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{3}{c}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{3}{a}\)
suy ra vt +\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
suy ra dpcm
dau = xay ra khi a=b=c
Lời giải:
Vì $a+b+c=1$ nên:
\(\text{VT}=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\right)\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{3}{4}\)
\(=\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}+\frac{1}{4}\left(\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\right)+\frac{3}{4}\)
\(=(\frac{ab}{a^2+b^2}+\frac{a^2+b^2}{4ab})+(\frac{bc}{b^2+c^2}+\frac{b^2+c^2}{4bc})+(\frac{ca}{c^2+a^2}+\frac{c^2+a^2}{4ac})+\frac{3}{4}\)
\(\geq 2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+2\sqrt{\frac{1}{4}}+\frac{3}{4}=\frac{15}{4}\) (áp dụng BĐT AM-GM)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\Leftrightarrow\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)
\(\Leftrightarrow\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}\le1\)
\(\Rightarrow1\ge\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\)
\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow\) đpcm
\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)
\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Với a, b, c > 0 có:
\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)
chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\)
theo BĐT cauchy schwars engel ta có
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)
\(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{\left(1+1\right)^2}{a+c}=\dfrac{4}{a+c}\)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1\right)^2}{b+c}=\dfrac{4}{b+c}\)
cộng vế theo vế ta có \(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{a+c}\right)\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{a+c}\)
vậy đpcm