Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2xy\sqrt{x}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2yz\sqrt{y}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2xz\sqrt{z}}=\dfrac{1}{xz}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\) ( 1 )

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2\sqrt{\dfrac{1}{x^2y^2}}=\dfrac{2}{xy}\\\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge2\sqrt{\dfrac{1}{y^2z^2}}=\dfrac{2}{yz}\\\dfrac{1}{z^2}+\dfrac{1}{x^2}\ge2\sqrt{\dfrac{1}{x^2z^2}}=\dfrac{2}{xz}\end{matrix}\right.\)

\(\Rightarrow2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)\)

\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)

\(\Leftrightarrow\dfrac{2\sqrt{x}}{x^3+y^2}+\dfrac{2\sqrt{y}}{y^3+z^2}+\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\) ( đpcm )

\(P=\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}\)

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

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a^2\left(b^2+3\right)}\le\dfrac{a^2+b^2+3}{2}\\\sqrt{b^2\left(c^2+3\right)}\le\dfrac{b^2+c^2+3}{2}\\\sqrt{c^2\left(a^2+3\right)}\le\dfrac{c^2+a^2+3}{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}\le\dfrac{2\left(a^2+b^2+c^2\right)+3}{2}=\dfrac{9}{2}\)

\(\Rightarrow\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\ge\dfrac{2\left(a^2+b^2+c^2\right)^2}{9}=2\)

Vì \(VT\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{\sqrt{a^2\left(b^2+3\right)}+\sqrt{b^2\left(c^2+3\right)}+\sqrt{c^2\left(a^2+3\right)}}\)

\(\Rightarrow VT\ge2\)

\(\Leftrightarrow\dfrac{a^3}{\sqrt{b^2+3}}+\dfrac{b^3}{\sqrt{c^2+3}}+\dfrac{c^3}{\sqrt{a^2+3}}\ge2\)

\(\Leftrightarrow P\ge2\)

Vậy \(P_{min}=2\)

b)

\(\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{ab}{a+b+2c}+\dfrac{bc}{2a+b+c}+\dfrac{ca}{a+2b+c}\le\dfrac{1}{4}\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab}{a+b+2c}=\dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\\\dfrac{bc}{2a+b+c}=\dfrac{bc}{a+b+a+c}\le\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{ca}{a+2b+c}=\dfrac{ca}{a+b+b+c}\le\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{ab}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)+\dfrac{bc}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{ca}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)

\(\Rightarrow VT\le\dfrac{ab}{4\left(a+c\right)}+\dfrac{ab}{4\left(b+c\right)}+\dfrac{bc}{4\left(a+b\right)}+\dfrac{bc}{4\left(a+c\right)}+\dfrac{ca}{4\left(a+b\right)}+\dfrac{ca}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\left[\dfrac{ab}{4\left(a+c\right)}+\dfrac{bc}{4\left(a+c\right)}\right]+\left[\dfrac{bc}{4\left(a+b\right)}+\dfrac{ca}{4\left(a+b\right)}\right]+\left[\dfrac{ca}{4\left(b+c\right)}+\dfrac{ab}{4\left(b+c\right)}\right]\)

\(\Rightarrow VT\le\dfrac{ab+bc}{4\left(a+c\right)}+\dfrac{bc+ca}{4\left(a+b\right)}+\dfrac{ca+ab}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\dfrac{b\left(a+c\right)}{4\left(a+c\right)}+\dfrac{c\left(a+b\right)}{4\left(a+b\right)}+\dfrac{a\left(b+c\right)}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\dfrac{a+b+c}{4}\)

\(\Rightarrow VT\le\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}\le\dfrac{1}{4}\) ( đpcm )

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

1)

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{1+a}\ge1-\dfrac{1}{1+b}-1-\dfrac{1}{1+c}=\dfrac{b}{1+b}+\dfrac{c}{1+c}\\\dfrac{1}{1+b}\ge1-\dfrac{1}{1+a}+1-\dfrac{1}{1+c}=\dfrac{a}{1+a}+\dfrac{c}{1+c}\\\dfrac{1}{1+c}\ge1-\dfrac{1}{1+a}+1-\dfrac{1}{1+b}=\dfrac{a}{1+a}+\dfrac{b}{1+b}\end{matrix}\right.\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{1+a}\ge\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\\\dfrac{1}{1+b}\ge\dfrac{a}{1+a}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{ac}{\left(1+a\right)\left(1+c\right)}}\\\dfrac{1}{1+c}\ge\dfrac{a}{1+a}+\dfrac{b}{1+b}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+b\right)}}\end{matrix}\right.\)

Nhân theo từng vế

\(\Rightarrow\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8\sqrt{\dfrac{a^2b^2c^2}{\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}\)

\(\Rightarrow\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow1\ge8abc\)

\(\Rightarrow abc\le\dfrac{1}{8}\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{2}\)

\(x^2-3mx-m=0\)

Theo định lý Viet

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=3m\\x_1x_2=-m\end{matrix}\right.\)

Ta có \(A=\dfrac{m^2}{x_2^2+3mx_1+3m}+\dfrac{x^2_1+3mx_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{x^2_2+\left(x_1+x_2\right)x_1+3m}+\dfrac{x^2_1+\left(x_1+x_2\right)x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{x^2_2+x^2_1+x_1x_2+3m}+\dfrac{x^2_1+x^2_2+x_1x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2+3m}+\dfrac{\left(x_1+x_2\right)^2-2x_1x_2+x_1x_2+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{\left(3m\right)^2-2\left(-m\right)+\left(-m\right)+3m}+\dfrac{\left(3m\right)^2-2\left(-m\right)+\left(-m\right)+3m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{9m^2+4m}+\dfrac{9m^2+4m}{m^2}\)

\(\Leftrightarrow A=\dfrac{m^2}{m\left(9m+4\right)}+\dfrac{m\left(9m+4\right)}{m^2}\)

\(\Leftrightarrow A=\dfrac{m}{9m+4}+\dfrac{9m+4}{m}\)

\(\Leftrightarrow A=\dfrac{m^2+\left(9m+4\right)^2}{\left(9m+4\right)m}\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow m^2+\left(9m+4\right)^2\ge2\sqrt{m^2\left(9m+4\right)^2}\)

\(\Rightarrow m^2+\left(9m+4\right)^2\ge2\left(9m+4\right)m\)

\(\Rightarrow\dfrac{m^2+\left(9m+4\right)^2}{\left(9m+4\right)m}\ge2\)

\(\Rightarrow A\ge2\)

Vậy \(A_{min}=2\)

17/15 x 4/11 + 8/15 x 4/11

= 4/11 x (17/15 + 8/15)

= 4/11 x 5/3

= 20/33

\(x^3+9x^2+11x-21=0\)

\(\Leftrightarrow x^3+7x^2+2x^2+11x-21=0\)

\(\Leftrightarrow x^2\left(x+7\right)+2x^2+11x-21=0\)

\(\Leftrightarrow x^2\left(x+7\right)+x^2+7x+x^2+4x-21=0\)

\(\Leftrightarrow x^2\left(x+7\right)+x\left(x+7\right)+x^2+4x-21=0\)

\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+4x-21=0\)

\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x^2+7x-3x-21=0\)

\(\Leftrightarrow\left(x+7\right)\left(x^2+x\right)+x\left(7+x\right)-3\left(7+x\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x^2+2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-3\\x=1\end{matrix}\right.\)

\(x^2-2\left(m+1\right)x+m+1=0\)

\(\Delta'=b'^2-ac\)

\(\Delta'=m^2+m\)

Để phương trình có 2 nghiệm phân biệt

\(\Rightarrow\Delta'>0\Leftrightarrow m^2+m>0\)

Lập bảng xét dấu

\(\Rightarrow\left[{}\begin{matrix}m< -1\\m>0\end{matrix}\right.\)

\(P=\dfrac{1}{a\left(2b+2c-1\right)}+\dfrac{1}{b\left(2c+2a-1\right)}+\dfrac{1}{c\left(2a+2b-1\right)}\)

\(P=\dfrac{1}{a\left[2b+2c-\left(a+b+c\right)\right]}+\dfrac{1}{b\left[2c+2a-\left(a+b+c\right)\right]}+\dfrac{1}{c\left[2a+2b-\left(a+b+c\right)\right]}\)

\(P=\dfrac{1}{a\left(b+c-a\right)}+\dfrac{1}{b\left(c+a-b\right)}+\dfrac{1}{c\left(a+b-c\right)}\)

\(P=\dfrac{1}{ab+ac-a^2}+\dfrac{1}{bc+ab-b^2}+\dfrac{1}{ca+bc-c^2}\)

Áp dụng bất đẳng thức cộng mẫu số

\(\Rightarrow P\ge\dfrac{\left(1+1+1\right)^2}{-a^2-b^2-c^2+2ab+2bc+2ca}=\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\) ( 1 )

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2-2\left(ab+bc+ca\right)\ge-\left(ab+bc+ca\right)\)

\(\Rightarrow-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]\le ab+bc+ca\)

\(\Rightarrow\dfrac{9}{-\left[a^2+b^2+c^2-2\left(ab+bc+ca\right)\right]}\ge\dfrac{9}{ab+bc+ca}\)

Từ ( 1 )

\(\Rightarrow P\ge\dfrac{9}{ab+bc+ca}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow1\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{1}{3}\ge ab+bc+ca\)

\(\Rightarrow27\le\dfrac{9}{ab+bc+ca}\)

\(\Rightarrow P\ge27\)

Vậy \(P_{min}=27\)