đề bài có đúng không

chắc chắn đúng

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

\(\Rightarrow\frac{1}{x+1}=2-\frac{1}{y+1}-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}=\frac{2\sqrt{yz}}{\sqrt{\left(y+1\right)\left(z+1\right)}}\)  (1)

(Vì x;y;z dương nên áp dụng BĐT Cô-si)

Chưng minh tương tự ta có: \(\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}=\frac{2\sqrt{xz}}{\sqrt{\left(x+1\right)\left(z+1\right)}}\) (2)

                                             \(\frac{1}{z+1}\ge\frac{2\sqrt{xy}}{\sqrt{\left(x+1\right)\left(y+1\right)}}\) (3)

Nhân (1) với (2) với (3) ta có:

giải tiếp

\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}=\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8\sqrt{\left(xyz\right)^2}}{\sqrt{\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2}}\)

Với x;y;z > 0 nên \(1\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Leftrightarrow1\ge8xyz\Leftrightarrow xyz\le\frac{1}{8}\)

Vậy ....

Ta có \(B=\frac{x^4}{x+xy}+\frac{y^4}{y+xy}\ge\frac{\left(x^2+y^2\right)^2}{x+y+2xy}\ge\frac{\left(x+y\right)^4}{4\left(x+y+2\right)}=\frac{a^4}{4\left(a+2\right)}\)

Ta có \(x+y\ge2\sqrt{xy}=2\Rightarrow a\ge2\)

Ta cần \(\frac{a^4}{4\left(a+2\right)}\ge1\Leftrightarrow a^4\ge4a+8\Leftrightarrow\frac{1}{2}a^4+\frac{1}{2}a^4\ge4a+8\)

Ta có\(\frac{1}{2}a^4\ge\frac{1}{2}.16=8;a^3\ge8\Rightarrow\frac{1}{2}a^4\ge4a\Rightarrow a^4\ge4a+8\)

=> B>=1

dấu = xảy ra <=> x=y=1

a)  ĐK:  \(0< a< 1\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

b)  Xét:  \(Q^3-Q=\left(a-1\right)^3-\left(a-1\right)=\left(a-1\right)^2\left(a-1-1\right)=\left(a-1\right)^2\left(a-2\right)\)

Do  \(a< 1\)=>  \(a-2< 0\) và   \(a-1< 0\) 

nên \(\left(a-1\right)^2\left(a-2\right)< 0\)

=>  \(Q^3-Q< 0\)

<=> \(Q^3< Q\)

Áp dụng BĐT Bu-nhi-a, ta có \(\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2\le3\left(2x+2y+2z\right)=6\)

=> A\(\le\sqrt{6}\)

dấu = xảy ra <=> x=y=z=1/3