bạn thử cosi xem

Áp dụng BĐT cosi:

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

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

Cộng VTV 3 BĐT trên:

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

Dấu \("="\Leftrightarrow x=y=z\)

Trả lời nhanh nha các bn, mik đang cần gấp, cảm ơn nhiều.

Kết hợp với giả thiết nêu ra ở đề bài, ta có vài biến đổi sau: 

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

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

Mặt khác, ta lại có: \(\left(x^2+x+1\right)\left(y^2+y+1\right)=x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1\)

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

Khi đó,  trừ đẳng thức  \(\left(1\right)\)  cho  đẳng thức  \(\left(2\right)\)  vế theo vế, ta được:

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

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

\(P=\left(x^2+y^2+2xy\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2+y^2+2xy}{x^2+y^2}\)

\(P=\left(x^2+y^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2xy\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+1+\dfrac{2xy}{x^2+y^2}\)

\(P\ge2xy.\dfrac{2}{xy}+\dfrac{2\left(x^2+y^2\right)}{xy}+1+\dfrac{2xy}{x^2+y^2}\)

\(P\ge\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}+\dfrac{3}{2}\left(\dfrac{x^2+y^2}{xy}\right)+5\)

\(P\ge2\sqrt{\dfrac{2xy\left(x^2+y^2\right)}{2xy\left(x^2+y^2\right)}}+\dfrac{3}{2}.\dfrac{2xy}{xy}+5=10\)

Dấu "=" xảy ra khi \(x=y\)