55.
\(3c^2\ge b^2+b^2+a^2\ge\dfrac{1}{3}\left(b+b+a\right)^2=\dfrac{1}{3}\left(2b+a\right)^2\)
\(\Rightarrow9c^2\ge\left(2b+a\right)^2\Rightarrow3c\ge2b+a\)
Do đó:
\(\dfrac{1}{a}+\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{4}{2b}\ge\dfrac{\left(1+2\right)^2}{a+2b}=\dfrac{9}{a+2b}\ge\dfrac{9}{3c}=\dfrac{3}{c}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
56.
\(\dfrac{x^2\left(y+z\right)}{yz}\ge\dfrac{4x^2\left(y+z\right)}{\left(y+z\right)^2}=\dfrac{4x^2}{y+z}\)
Tương tự:
\(\dfrac{y^2\left(z+x\right)}{zx}\ge\dfrac{4y^2}{z+x}\) ; \(\dfrac{z^2\left(x+y\right)}{xy}\ge\dfrac{4z^2}{x+y}\)
Cộng vế với vế:
\(P\ge\dfrac{4x^2}{y+z}+\dfrac{4y^2}{z+x}+\dfrac{4z^2}{x+y}\ge\dfrac{4\left(x+y+z\right)^2}{2\left(x+y+z\right)}=2\left(x+y+z\right)=2\)
Vậy \(P_{min}=2\) khi \(x=y=z=\dfrac{1}{3}\)
57.
\(\dfrac{a^2}{b+c-a}+\dfrac{b^2}{c+a-b}+\dfrac{c^2}{a+b-c}\ge\dfrac{\left(a+b+c\right)^2}{b+c-a+c+a-b+a+b-c}=\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)
Dấu "=" xảy ra khi \(a=b=c\)
58.
\(VT=\dfrac{a\left(1+c\right)+b\left(1+a\right)+c\left(1+b\right)}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}=\dfrac{ab+bc+ca+a+b+c}{1+a+b+c+ab+bc+ca+abc}\)
\(VT\ge\dfrac{ab+bc+ca+a+b+c}{a+b+c+ab+bc+ca+2}\)
Nên ta chỉ cần chứng minh:
\(\dfrac{ab+bc+ca+a+b+c}{a+b+c+ab+bc+ca+2}\ge\dfrac{3}{4}\)
\(\Leftrightarrow a+b+c+ab+bc+ca\ge6\)
Điều này hiển nhiên đúng do: \(a+b+c+ab+bc+ca\ge6\sqrt[6]{\left(abc\right)^3}=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
59.
\(\dfrac{a^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{1+a}{8}+\dfrac{1+b}{8}\ge3\sqrt[3]{\dfrac{a^3\left(1+a\right)\left(1+b\right)}{64\left(a+1\right)\left(1+b\right)}}=\dfrac{3a}{4}\)
Tương tự:
\(\dfrac{b^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge\dfrac{3b}{4}\)
\(\dfrac{c^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{1+c}{8}+\dfrac{1+a}{8}\ge\dfrac{3c}{4}\)
Cộng vế với vế:
\(VT+\dfrac{3}{4}+\dfrac{a+b+c}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{abc}}{2}-\dfrac{3}{4}=\dfrac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
