Đề thi HSG Toán 9 Huyện Hoàng mia năm 2019-2020 đó 

ko dùng điều kiện :) 

\(sigma\sqrt{\frac{1+a^2}{b+c}}\ge sigma\frac{a+1}{\sqrt{2\left(b+c\right)}}\ge sigma\frac{2\left(a+1\right)}{b+c+2}=sigma\left(\frac{2a^2}{ab+ca+2a}+\frac{2}{b+c+2}\right)\)

\(\ge\frac{2\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)+2\left(a+b+c\right)}+\frac{18}{2\left(a+b+c\right)+6}\)

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

"=" \(\Leftrightarrow\)\(a=b=c=1\)

Áp dụng BĐT Cauchy ta có:

\(\left(a+b-c\right)\left(b+c-a\right)\le\left(\frac{a+b-c+b+c-a}{2}\right)^2=b^2\)Dấu "=" xảy ra khi\(a+b-c=b+c-a\Leftrightarrow a=c\)

\(\left(b+c-a\right)\left(c+a-b\right)\le\left(\frac{b+c-a+c+a-b}{2}\right)^2=c^2\)Dấu "=" xảy ra khi\(b+c-a=c+a-b\Leftrightarrow a=b\)

\(\left(c+a-b\right)\left(a+b-c\right)\le\left(\frac{c+a-b+a+b-c}{2}\right)^2=a^2\)Dấu "=" xảy ra khi\(c+a-b=a+b-c\Leftrightarrow b=c\)

Nhân vế theo vế các BĐT trên, ta có:

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

\(\Rightarrowđfcm\)Dấu "=" khi a=b=c.

HPT \(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{2\left(x^2+y^2\right)}+2\sqrt{xy}=16\\x+y+2\sqrt{xy}=16\end{cases}}\)

Như vậy ta có: \(\sqrt{2\left(x^2+y^2\right)}=x+y\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)

Bí.

Áp dụng bất đẳng thức Bunyakovsky 

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

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

Tương tự ta có : \(\hept{\begin{cases}\frac{y^2}{y^4+xz}\le\frac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\frac{z^2}{z^4+xy}\le\frac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{cases}}\)

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

Chứng minh rằng :

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

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

Áp dụng bất đẳng thức Cauchy

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

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

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

Tương tự ta có : \(\hept{\begin{cases}\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\frac{1}{4\sqrt{xz}}\\\frac{z^2}{\left(z^2+\sqrt{zy}\right)^2}\le\frac{1}{4\sqrt{xy}}\end{cases}}\)

\(\Leftrightarrow\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\)

\(\le\frac{1}{4}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{xz}\right)\)

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

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

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

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

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

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

Áp dụng bất đẳng thức Cauchy

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

Tương tự ta có : \(\hept{\begin{cases}\frac{1}{\sqrt{xz}}\le\frac{\frac{1}{x}+\frac{1}{z}}{2}\\\frac{1}{\sqrt{xy}}\le\frac{\frac{1}{z}+\frac{1}{y}}{2}\end{cases}}\)

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

Áp dụng bất đẳng thức Cauchy 

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

Tương tự ta có :

\(\hept{\begin{cases}\frac{y}{xz}+\frac{z}{xy}\ge\frac{2}{x}\\\frac{x}{zy}+\frac{z}{xy}\ge\frac{2}{y}\end{cases}}\)

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

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

Từ (1) và (2) 

\(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le3\left(đpcm\right)\)

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

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

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

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

Dấu "=" xảy ra khi \(x=y=z=1\)

Chúc bạn học tốt !!!

\(\text{Σ}\frac{x^2}{x^4+yz}\le\text{Σ}\frac{x^2}{2x^2\sqrt{yz}}=\text{Σ}\frac{1}{2\sqrt{yz}}\le\text{Σ}\frac{\frac{1}{y}+\frac{1}{z}}{4}=\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{2}=\frac{\frac{xy+yz+xz}{xyz}}{2}=\frac{\frac{3\left(xy+yz+xz\right)}{x^2+y^2+z^2}}{2}\)(1)

Dễ dàng CM được: \(x^2+y^2+z^2\ge xy+yz+xz\)

Thay vào (1) -> dpcm