Ờ thì AM-GM (là Cô si ko âm đây)

\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}\cdot\frac{y+z}{4}}=2\cdot\frac{x}{2}=x\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y;\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\)

Cộng theo vế 3 BĐT trên ta có:

\(P+\frac{2\left(x+y+z\right)}{4}\ge x+y+z\Leftrightarrow P\ge1\)

ĐẲng thức xảy ra khi \(x=y=z=\frac{2}{3}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dưới dạng Engel ta có :

\(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}=\frac{2}{2}=1\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{2}{3}\)

Vậy \(P_{min}=1\) tại \(x=y=z=\frac{2}{3}\)

Đinh Đức Hùng ơi, cái cauchy-schwars dưới dạng engel mình chưa học, mới học cái bđt cauchy a+b >= căn ab với a,b ko âm thoy à

Từ giả thiết ta có :

\(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)

ta có : \(Q=\frac{y+2}{x^2}+\frac{z+2}{y^2}+\frac{x+2}{z^2}\)

\(=\frac{\left(x+1\right)+\left(y+1\right)}{x^2}+\frac{\left(y+1\right)+\left(z+1\right)}{y^2}+\frac{\left(z+1\right)+\left(x+1\right)}{z^2}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\left(x+1\right)\left(\frac{1}{z^2}+\frac{1}{x^2}\right)+\left(y+1\right)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)+\left(z+1\right)\left(\frac{1}{y^2}+\frac{1}{z^2}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

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

\(=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2\)

Áp dụng bđt \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Dấu " = " xảy ra khi và chỉ khi a = b = c

Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\ge3\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=3\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\sqrt{3}\)

Do đó : \(Q\ge\sqrt{3}+2\). Dấu " = " xảy ra 

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\\z+y+z=xyz\end{cases}\Leftrightarrow x=y=z=\sqrt{3}}\)

Vậy Min \(Q=\sqrt{3}+2\)khi \(x=y=z=\sqrt{3}\)

Ta có: \(\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2yx}+\frac{z^4}{zx+2zy}\)

Áp dụng BĐT Cauchy Schwarz, ta có:

\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2yx}+\frac{z^4}{zx+2zy}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)

=> ĐPCM

Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{\sqrt{3}}\)

Áp dụng BĐT Cosi cho 2 số dương, ta có:

\(\frac{9x^3}{y+2z}+x\left(y+2z\right)\ge6x^2;\frac{9y^3}{z+2x}+y\left(z+2x\right)\ge6y^2;\frac{9z^3}{x+2y}+z\left(x+2y\right)\ge6z^3\)

Lại có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)

Do đó \(\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}+3\left(xy+yz+zx\right)\ge6\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\frac{9x^3}{y+2z}+\frac{9y^3}{z+2x}+\frac{9z^3}{x+2y}\ge6\left(x^2+y^2+z^2\right)-3\left(xy+yz+zx\right)\ge3\left(x^2+y^2+z^2\right)\)

\(\Leftrightarrow\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\ge\frac{x^2+y^2+z^2}{3}=\frac{1}{3}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{\sqrt{3}}\)

1/ Đầu tiên ta chứng minh: \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\) (1)

\(\Leftrightarrow\Sigma_{cyc}\left(\frac{a^3}{b}-a^2\right)\ge0\Leftrightarrow\Sigma_{cyc}\left(\frac{a^2\left(a-b\right)}{b}-a\left(a-b\right)\right)+\Sigma_{cyc}a\left(a-b\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{a\left(a-b\right)^2}{b}+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{a\left(a-b\right)^2}{b}+\Sigma_{cyc}\frac{1}{2}\left(a-b\right)^2\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{\left(a-b\right)^2\left(2a+b\right)}{2b}\ge0\)

BĐT cuối đúng nên (1) đúng. (*)

Bây giờ ta đi chứng minh: \(a^2+b^2+c^2\ge5\)

Đặt \(\left(a+b+c;ab+bc+ca\right)\rightarrow\left(3u;3v^2\right)\) thì \(3u=9-3v^2\)

và \(a^2+b^2+c^2=\left(3u\right)^2-6v^2=\left(9-3v^2\right)^2-6v^2\)

\(=\left(3v^2-9\right)^2-6v^2=9v^4-60v^2+81\)

Đặt \(v^2=t\ge0\) .Ta cần tìm min của: \(9t^2-60t+81\)

Ta có: \(9t^2-60t+81=\left(3t-10\right)^2-19\ge-19\)

Dấu "=" xảy ra khi t = 10/3 tức là v= \(\sqrt{\frac{10}{3}}\)....

Em thấy có gì đó sai sai thì phải ạ:((

Câu 1:

\(\frac{a^3}{b}+ab\ge2a^2\) ; \(\frac{b^3}{c}+bc\ge2b^2\); \(\frac{c^3}{a}+ac\ge2c^2\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}+ab+ac+bc\ge2\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+ac+bc\right)\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(a^2+b^2+c^2\right)=a^2+b^2+c^2\)

//

\(a+b+c+ab+ac+bc\le a+b+c+\frac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{3}+\left(a+b+c\right)\ge9\)

\(\Rightarrow\left(a+b+c-\frac{3\sqrt{13}-3}{2}\right)\left(a+b+c+\frac{3\sqrt{13}+3}{2}\right)\ge0\)

\(\Rightarrow a+b+c\ge\frac{3\sqrt{13}-3}{2}\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\ge\frac{1}{3}\left(\frac{3\sqrt{13}-3}{2}\right)^2=\frac{21-3\sqrt{13}}{2}>5\)

\(\Rightarrow a^2+b^2+c^2>5\)

Dấu "=" ko xảy ra

Bài 2:

Ta có: \(VT^2=\left(x+\sqrt{2-x}\right)^2\le2\left(x^2+2-x^2\right)=4\)

\(\Rightarrow VT\le2\)

\(VP=4y^2+4y+1+2=\left(2y+1\right)^2+2\ge2\)

\(\Rightarrow VT\le VP\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}2y+1=0\\x=\sqrt{2-x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-\frac{1}{2}\end{matrix}\right.\)

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

\(\frac{x^2+y^2+4}{x+y}=\frac{z\left(x+y\right)+4}{x+y}=z-x-y+\frac{4}{x+y}+x+y\ge z-x-y+4\)

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

\(1=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{z}\right)+\frac{1}{2}\left(\frac{y}{z}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{z}{x}+\frac{x}{y}\right)\)

\(\ge\sqrt{\frac{x}{y}.\frac{y}{z}}+\sqrt{\frac{y}{z}.\frac{z}{x}}+\sqrt{\frac{z}{x}.\frac{x}{y}}=VP\) (rút gọn lại thôi:v)

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)

Áp dụng bđt AM-GM ta có:

\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=x\)

\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}=y\)

\(\frac{z^2}{y+x}+\frac{y+x}{4}\ge2\sqrt{\frac{z^2}{y+x}.\frac{y+x}{4}}=z\)

Cộng các vế của các bđt trên ta được:

\(P+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow P\ge\frac{x+y+z}{2}=1\)

Dấu"="Xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)

Áp dụng Svac - xơ:

\(P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{2^2}{2.2}=1\)

(Dấu "="\(\Leftrightarrow x=y=z=\frac{2}{3}\))

Ta có x√(1-y²)<= (x² + 1 - y²)/2

y√(1-z²)<=  (y² +1 - z²)/2

z√(1- x²)<= (z² + 1 - x²)/2

=>x√(1-y²) +y√(1-z²)z+√(1- x²)<=3/2

Đấu đẳng thức xảy ra khi: x² = 1 - y²

y² = 1-z²

z²  = 1- x²

Cộng vế theo vế ta được điều phải chứng minh

Thanks nhiều