\(a^3+a\ge2a^2\);\(b^3+b\ge2b^2\)

\(a^3+b^3\ge2\left(a^2+b^2\right)-\left(a+b\right)\ge4-\sqrt{2\left(a^2+b^2\right)}=4-2=2\)

\(2=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow a+b\le2\)

\(\left(a^3+b^3\right)\left(a+b\right)\ge\left(a^2+b^2\right)^2=4\)

\(a^3+b^3\ge\dfrac{4}{a+b}\ge\dfrac{4}{2}=2\)

\(\dfrac{x^4}{a}+\dfrac{y^4}{b}\ge\dfrac{\left(x^2+y^2\right)^2}{a+b}=\dfrac{1}{a+b}\)(cauchy-schwarz)

dấu = xảy ra khi \(\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow bx^2=ay^2\)

31^11<32^11=(2^5)^11=2^55

17^14>16^14=(2^4)^14=2^56

suy ra:17^14>2^56>2^55>31^11

17^14>31^11

đề phải ntn chứ \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\)

Áp dụng BĐT cauchy ta có:\(\left\{{}\begin{matrix}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\x^2+z^2\ge2xz\end{matrix}\right.\)

\(P\le\dfrac{1}{4xy+4x+4}+\dfrac{1}{4yz+4y+4}+\dfrac{1}{4xz+4z+4}=\dfrac{1}{4}\left(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+x+1}\right)\)

xét biểu thức \(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{zx+z+1}=\dfrac{1}{xy+x+1}+\dfrac{x}{1+yx+x}+\dfrac{xy}{x+1+xy}=\dfrac{xy+x+1}{xy+x+1}=1\)do đó \(P\le\dfrac{1}{4}\)

dấu = xảy ra khi x=y=z=1

\(\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}=\sqrt{\left(1+\frac{1}{a}-\frac{1}{a+1}\right)^2}=\left|1+\frac{1}{a}-\frac{1}{a+1}\right|\)

tao chửi tiếp tao chửi tiếp tao chửi chưa đã đâu olm chó đẻ tao chửi chưa có đã mày nghe chưa 

2 Pt đầu khử a ,2 pt sau khử a ,ta được HPT 2 ẩn b,c