Xét hiệu : \(\frac{x^4+y^4}{\left(xy\right)^2}-\frac{x^2+y^2}{ab}\)

\(\Leftrightarrow\frac{\left(x^4+y^4\right)-\left(x^3y+yx^3\right)}{\left(xy\right)^2}\)

\(\Leftrightarrow\frac{x^3\left(x-y\right)+y^3\left(y-x\right)}{\left(xy\right)^2}\)

\(\Leftrightarrow\frac{\left(x-y\right)^2\left(x^2+xy+y^2\right)}{\left(xy\right)^2}\ge0\forall x,y\)

=> đpcm

Đề có sai hay thiếu gì k bạn, có đk x,y >0 hay k ?

\(BĐT\Leftrightarrow\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\right)-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}-2\right)\ge0\) (Luôn đúng vì \(\frac{x}{y}+\frac{y}{x}\ge2\forall x;y>0\))

Đặt \(\dfrac{x}{y}+\dfrac{y}{x}=a\)\(\Rightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2=a^2\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}=a^2-2\)

Ta có \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4=a^2-2+4=a^2+2\)

\(3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)=3a\)

Ta có \(a^2+2-3a=a^2-2.a.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{1}{4}=\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\)

lạ có \(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2}{xy}-\dfrac{2xy}{xy}+\dfrac{y^2}{xy}=\dfrac{\left(x-y\right)^2}{xy}\ge0\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}\ge2\)\(\Rightarrow a\ge2\Rightarrow a-\dfrac{3}{2}\ge\dfrac{1}{2}\)\(\Rightarrow\left(a-\dfrac{3}{2}\right)^2\ge\dfrac{1}{4}\Rightarrow\left(a-\dfrac{3}{2}\right)^2-\dfrac{1}{4}\ge0\)

\(\Rightarrow a^2+2-3a\ge0\Rightarrow a^2+2\ge3a\Rightarrow\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

\(\left\{{}\begin{matrix}x;y>0\\\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)

từ (2) có \(\Leftrightarrow\left(\dfrac{x^2}{y^2}+2.\dfrac{x}{y}.\dfrac{y}{x}+\dfrac{y^2}{x^2}\right)+2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\right]-\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)-2\right]\ge0\)

\(\Leftrightarrow\left(\dfrac{x}{y}+\dfrac{y}{x}-2\right)\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)\ge0\) (3)

từ (1) có \(\dfrac{x}{y}+\dfrac{y}{x}=\left(\sqrt{\dfrac{x}{y}}-\sqrt{\dfrac{y}{x}}\right)^2+2\ge2\) (4)

từ (4) ; \(\left\{{}\begin{matrix}\left(\dfrac{x}{y}+\dfrac{y}{x}-1\right)>0\\\dfrac{x}{y}+\dfrac{y}{x}-2\ge0\end{matrix}\right.\) (I)

từ (I) => (3) đúng mọi phép biến đổi là <=> đẳng thức khi \(\dfrac{x}{y}=\dfrac{y}{x}\Rightarrow x=y\)=> dpcm

ta có: \(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+2=\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2\)(1)

vì x,y >0

nên \(\dfrac{x}{y}+\dfrac{y}{x}=t,t\ge2\)ta được:

\(\left(\dfrac{x}{y}+\dfrac{y}{x}\right)^2+2\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

<=>\(t^2+2\ge3t< =>t^2+2-3t\ge0< =>t^2-t-2t+2\ge0< =>t\left(t-1\right)-2\left(t-1\right)\ge0< =>\left(t-1\right)\left(t-2\right)\ge0\)

(BĐT cuối luôn đúng vì t > hoặc = 0) (2)

từ (1) và (2)=>\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}+4\ge3\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

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

Dấu "=" xảy ra khi \(x=y=z\)

Hoặc:

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

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

Cộng vế với vế ta có đpcm

#Bạn_về_tìm_hiểu_các_BĐT_cơ_bản_như_AM-GM_hay_Cauchy-Schwarz_nhé.

Áp dụng BĐT Cauchy-Schwarz dạng phân thức:

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

\(''=''\Leftrightarrow x=y\)

Ta có:

\(\frac{x^2}{y^2}+\frac{y^2}{x^2}+4\ge3\left(\frac{x}{y}+\frac{y}{x}\right)\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-2+4-3\left(\frac{x}{y}+\frac{y}{x}\right)\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-3\left(\frac{x}{y}+\frac{y}{x}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+1\right)-3\left(\frac{x}{y}+\frac{y}{x}-1\right)\ge0\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}-1\right)\left(\frac{x}{y}+\frac{y}{x}+2\right)\ge0\left(1\right)\)

Đến đây có 2 cách giải quyết

Cách 1:

\(\left(1\right)\Leftrightarrow\frac{x^2-xy+y^2}{xy}\cdot\frac{\left(x+y\right)^2}{xy}\ge0\)

\(\Leftrightarrow\frac{\left(x+y\right)^2\left(x^2-xy+y^2\right)}{x^2y^2}\ge0\)

\(\Leftrightarrow\frac{\left(x+y\right)^2\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{x^2y^2}\ge0\left(true!!!\right)\)

Cách 2 là đặt ẩn:)

Đặt \(\frac{x}{y}+\frac{y}{x}=t\Rightarrow t^2=\left(\frac{x}{y}+\frac{y}{x}\right)^2\ge4\cdot\frac{x}{y}\cdot\frac{y}{x}=4\)

\(\Rightarrow\left|t\right|\ge2\)

Khi đó ta có:

\(\left(t+1\right)\left(t-2\right)\ge0\)

Nếu \(t\ge2\Rightarrow t+1>0;t-2\ge0\Rightarrow\left(t+1\right)\left(t-2\right)\ge0\)

Nếu \(t\le-2\Rightarrow t+1< 0;t-2< 0\Rightarrow\left(t+1\right)\left(t-2\right)>0\)

=> đpcm

coolkid cách 1 viết sai rồi nha Cool kid, phải là:

\(VT-VP=\frac{\left(x-y\right)^2\left[\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]}{x^2y^2}\ge0\) (chú ý là (x - y)² chứ ko phải (x + y)² nha!)

Biến đổi \(\frac{x}{y^3-1}-\frac{y}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}=\frac{\left(x^4-y^4\right)-\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)

(Do x+y=1 => \(\hept{\begin{cases}y-1=-x\\x-1=-y\end{cases}}\))

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

\(=\frac{\left(x-y\right)\left(x^3+y^3-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)

\(=\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)

\(=\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}=\frac{-2\left(x-y\right)}{x^2y^2+3}\)

\(\Rightarrow\frac{x}{y^3-1}-\frac{y}{x^3-1}+\frac{2\left(x-y\right)}{x^2y^2+3}=0\left(đpcm\right)\)