\(P=\dfrac{a^3}{b^2+ab+bc+ca}+\dfrac{b^3}{c^2+ab+bc+ca}+\dfrac{c^3}{a^2+ab+bc+ca}=\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}\)

Ta có:

\(\dfrac{a^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{a+b}{8}+\dfrac{b+c}{8}\ge\dfrac{3a}{4}\)

\(\dfrac{b^3}{\left(a+c\right)\left(b+c\right)}+\dfrac{a+c}{8}+\dfrac{b+c}{8}\ge\dfrac{3b}{4}\)

\(\dfrac{c^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge\dfrac{3c}{4}\)

Cộng vế:

\(P+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow P\ge\dfrac{1}{4}\left(a+b+c\right)\ge\dfrac{1}{4}.\sqrt{3\left(ab+bc+ca\right)}=\dfrac{\sqrt{3}}{4}\)

Ta có: \(1=a^2+b^2+c^2\ge ab+bc+ca\).

\(P=\dfrac{a^3}{b+2c}+\dfrac{b^3}{c+2a}+\dfrac{c^3}{a+2b}=\dfrac{a^4}{ab+2ca}+\dfrac{b^4}{bc+2ab}+\dfrac{c^4}{ca+2bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3\left(ab+bc+ca\right)}=\dfrac{1}{3\left(ab+bc+ca\right)}\ge\dfrac{1}{3}\)

Dấu \(=\) xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\).

\(2.\) Ba số dương a,b,c chứ?

câu 1 bn bình phương vế 2x+y đi nhé!

Áp dụng bất đẳng thức Bunyakovsky cho hai bộ số  \(\left[\left(\sqrt{2}\right)^2+1\right]\) và  \(\left(2x^2+y^2\right)\), ta được:

\(\left[\left(\sqrt{2}\right)^2+1^2\right]\left(2x^2+y^2\right)\ge\left(\sqrt{2}.\sqrt{2}x+1.y\right)^2\)

\(\Rightarrow\)  \(3\left(2x^2+y^2\right)\ge\left(2x+y\right)^2=3^2=9\)

\(\Rightarrow\)  \(2x^2+y^2\ge3\)

Dấu  \("="\)  xảy ra  \(\Leftrightarrow\)  \(\frac{\sqrt{2}}{\sqrt{2}x}=\frac{1}{y}\)  \(\Leftrightarrow\)  \(x=y=1\)

a)A=x(x+1)(x+2)(x+3)

\(=\left(x^2+3x\right)\left(x^2+3x+2\right)\)

Đặt \(t=x^2+3x\) ta đc:

\(t\left(t+2\right)\)\(=t^2+2t+1-1\)

\(=\left(t+1\right)^2-1\ge-1\)

Dấu = khi \(t=-1\Rightarrow x^2+3x=-1\)\(\Rightarrow\)\(x=\frac{-3\pm\sqrt{5}}{2}\)

Vậy MinA=-1 khi \(x=\frac{-3\pm\sqrt{5}}{2}\)

b)\(B=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Với a,b,c dương ta áp dụng Bđt Cô si 3 số:

\(a+b+c\ge3\sqrt[3]{abc}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Dấu = khi a=b=c

Vậy MinB=9 khi a=b=c

c)\(C=a^2+b^2+c^2\)

Áp dụng Bđt Bunhiacopski 3 cặp số ta có:

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(1a+1b+1c\right)^2=\left(\frac{3}{2}\right)^2=\frac{9}{4}\)

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\frac{9}{4}\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{3}{4}\)

\(\Rightarrow C\ge\frac{3}{4}\)

Dấu = khi \(a=b=c=\frac{1}{2}\)

Vậy MinC=\(\frac{3}{4}\) khi \(a=b=c=\frac{1}{2}\)