\(BĐT\Leftrightarrow35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+8\left(a^2+b^2+c^2\right)\)

Cần chứng minh rằng \(8\left(a^2+b^2+c^2\right)\ge\dfrac{72abc}{a+b+c}\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge9abc\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}=9abc\left(đpcm\right)\)

Vậy \(8\left(a^2+b^2+c^2\right)\ge\dfrac{72abc}{a+b+c}\) \(\Rightarrow9\left(a+b+c\right)^2+8\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\)

\(\Rightarrow\)\(35\left(a^2+b^2+c^2\right)\ge9\left(a+b+c\right)^2+\dfrac{72abc}{a+b+c}\) ( đpcm )

Chứng minh bổ đề: \(\dfrac{4x}{3-4x^2}\ge4x^2\)

\(\Leftrightarrow1+4x^3\ge3x\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{2}+4x^3\ge3x\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{2}+4x^3\ge3\sqrt[3]{\dfrac{4x^3}{4}}=3x\left(đpcm\right)\)

Áp dụng bổ đề cho các phân thức còn lại và thu lại ta có

\(P\ge4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+xz\right)=3\)

Vậy \(P_{min}=3\)

Ta có \(\dfrac{ab+c}{c+1}=\dfrac{ab+c\left(a+b+c\right)}{\left(a+c\right)+\left(b+c\right)}=\dfrac{\left(a+c\right)\left(b+c\right)}{\left(a+c\right)+\left(b+c\right)}\)

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

Đặt \(\left\{{}\begin{matrix}a+c=x\\b+c=y\\a+b=z\end{matrix}\right.\) \(\Rightarrow x+y+z=2\)

\(\Rightarrow VT\Leftrightarrow\dfrac{xy}{x+y}+\dfrac{yz}{z+y}+\dfrac{xz}{x+z}\)

Áp dụng bất đẳng thức \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)

\(\Rightarrow\dfrac{xy}{x+y}\le\dfrac{xy}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{y}{4}+\dfrac{x}{4}\)

Thiết lập tương tự và thu lại ta có

\(\Rightarrow VT\le\dfrac{2\left(x+y+z\right)}{4}=1\) ( đpcm )

\(\Leftrightarrow\dfrac{ab+c}{c+1}+\dfrac{bc+a}{a+1}+\dfrac{ac+b}{b+1}\le1\)

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

\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)

Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)

\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)

Ta có \(abc\ge1\)

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

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)

\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)

\(\Sigma\left(\dfrac{x^5-x^2}{x^5+y^2+z^2}\right)\ge0\)

\(\Leftrightarrow\Sigma\left(1-\dfrac{x^5-x^2}{x^5+y^2+z^2}\right)\le3\)

\(\Leftrightarrow\Sigma\left(\dfrac{x^2+y^2+z^2}{x^5+y^2+z^2}\right)\le3\)

\(\Leftrightarrow\dfrac{1}{x^5+y^2+z^2}+\dfrac{1}{y^5+x^2+z^2}+\dfrac{1}{z^5+x^2+y^2}\le\dfrac{3}{x^2+y^2+z^2}\)

Áp dụng bất đẳng thức Bunyakovsky

\(\Rightarrow\left(x^5+y^2+z^2\right)\left(\dfrac{1}{x}+y^2+z^2\right)\ge\left(x^2+y^2+z^2\right)^2\)

\(\Rightarrow\dfrac{1}{x^5+y^2+z^2}\le\dfrac{\dfrac{1}{x}+y^2+z^2}{\left(x^2+y^2+z^2\right)^2}\)

Thiết lập tương tự và thu lại ta có

\(\Rightarrow\dfrac{1}{x^5+y^2+z^2}+\dfrac{1}{y^5+x^2+z^2}+\dfrac{1}{z^5+x^2+y^2}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}\)

Chứng minh rằng \(\dfrac{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}\le\dfrac{3}{x^2+y^2+z^2}\)

\(\Leftrightarrow\dfrac{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}\le\dfrac{x^2+y^2+z^2+2\left(x^2+y^2+z^2\right)}{\left(x^2+y^2+z^2\right)^2}\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le x^2+y^2+z^2\) ( vì \(xyz=1\) )

\(\Leftrightarrow xy+yz+xz\le x^2+y^2+z^2\) ( luôn đúng theo hệ quả của bất đẳng thức Cauchy )

\(\Rightarrow\) đpcm

Dấu " = " xảy ra khi \(x=y=z=1\)

\(P=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

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

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow2\sqrt{\left(a+b\right)\left(b+c\right)}\le a+2b+c\)

Tượng tự ta có \(\left\{{}\begin{matrix}2\sqrt{\left(b+c\right)\left(c+a\right)}\le a+b+2c\\2\sqrt{\left(a+b\right)\left(a+c\right)}\le2a+b+c\end{matrix}\right.\)

\(\Rightarrow P^2\le2+4\left(a+b+c\right)\)

\(\Rightarrow P\le\sqrt{6}\)

Vậy \(P_{max}=\sqrt{6}\)

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

có số chữ số là: (99 - 10):1 + 1 = 90 (số có 2 chứ số )

mà có 9 chữ số giống nhau

=> số chữ số khác nhau có 2 chữ số là : 90 - 9 = 81 số

a) \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{x\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

\(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x^3}+1}+\dfrac{2}{x-\sqrt{x}+1}\)

\(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\dfrac{2}{x-\sqrt{x}+1}\)

\(A=\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}+1\right)}{\left(x-\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\)

\(A=\dfrac{x-\sqrt{x}+1-3+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

b) Chứng minh \(A\ge0\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(A=\dfrac{\sqrt{x}}{\sqrt{x^2}-2\sqrt{x}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}+1}=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Mà \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) và \(\sqrt{x}\ge0\)

\(\Rightarrow A=\dfrac{\sqrt{x}}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\ge0\) (1)

Chứng minh \(A\le1\)

Ta có \(A=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\le1\)

\(\Leftrightarrow\sqrt{x}\le x-\sqrt{x}+1\)

\(\Leftrightarrow2\sqrt{x}\le x+1\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow x+1\ge2\sqrt{x}\) ( luôn đúng với mọi \(x\ge0\) )

Vậy \(A\le1\) (2)

Từ (1) và (2)

\(\Rightarrow0\le A\le1\) ( đpcm )

4) Áp dụng bất đẳng thức Bunyakovsky

\(\Rightarrow\left(x^4+yz\right)\left(1+1\right)\ge\left(x^2+\sqrt{yz}\right)^2\)

\(\Rightarrow\dfrac{x^2}{x^4+yz}\le\dfrac{2x^2}{\left(x^2+\sqrt{yz}\right)^2}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{y^4+xz}\le\dfrac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\dfrac{z^2}{z^4+xy}\le\dfrac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{matrix}\right.\)

\(\Rightarrow VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)

Chứng minh rằng \(2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow x^2+\sqrt{yz}\ge2\sqrt{x^2\sqrt{yz}}=2x\sqrt{\sqrt{yz}}\)

\(\Rightarrow\left(x^2+\sqrt{yz}\right)^2\ge4x^2\sqrt{yz}\)

\(\Rightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}\le\dfrac{x^2}{4x^2\sqrt{yz}}=\dfrac{1}{4\sqrt{yz}}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\dfrac{1}{4\sqrt{xz}}\\\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4\sqrt{xy}}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)

Theo đề bài ta có \(x^2+y^2+z^2=3xyz\)

\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=3\)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)

\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{1}{\sqrt{xy}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{2}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{xz}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{z}}{2}\\\dfrac{1}{\sqrt{yz}}\le\dfrac{\dfrac{1}{z}+\dfrac{1}{y}}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (1)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}\ge2\sqrt{\dfrac{1}{z^2}}=\dfrac{2}{z}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{2}{x}\\\dfrac{x}{zy}+\dfrac{z}{xy}\ge\dfrac{2}{y}\end{matrix}\right.\)

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

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

Từ (1) và (2)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\) ( đpcm )

Vậy \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)

\(\Rightarrow2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)

Mà \(VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)

\(\Rightarrow VT\le\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=1\)