\(P=\dfrac{1}{3x\left(y+z\right)+x+y+z}+\dfrac{1}{3y\left(z+x\right)+x+y+z}+\dfrac{1}{3z\left(x+y\right)+x+y+z}\)

\(P\le\dfrac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3y\left(z+x\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3z\left(x+y\right)+3\sqrt[3]{xyz}}\)

\(P\le\dfrac{1}{3x\left(y+z\right)+3}+\dfrac{1}{3y\left(z+x\right)+3}+\dfrac{1}{3z\left(x+y\right)+3}\)

Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3\left(b^3+c^3\right)+1}+\dfrac{1}{b^3\left(c^3+a^3\right)+1}+\dfrac{1}{c^3\left(a^3+b^3\right)+1}\right)\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3bc\left(b+c\right)+1}+\dfrac{1}{b^3ac\left(a+c\right)+1}+\dfrac{1}{c^3ab\left(a+b\right)+1}\right)\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{bc}{a\left(b+c\right)+bc}+\dfrac{ac}{b\left(a+c\right)+ac}+\dfrac{ab}{c\left(a+b\right)+ab}\right)=\dfrac{1}{3}\)

\(P_{max}=\dfrac{1}{3}\) khi \(a=b=c=1\) hay \(x=y=z=1\)

Do \(x;y\in\left[0;2\right]\Rightarrow\left\{{}\begin{matrix}x\left(2-x\right)\ge0\\y\left(2-y\right)\ge0\end{matrix}\right.\) \(\Rightarrow2x^2+4y^2\le4x+8y\)

\(P\le3^0+5^0+3^z+4\left(x+2y\right)=2+3^z+4\left(6-z\right)=3^z-4z+26\)

Xét hàm \(f\left(z\right)=3^z-4z+26\) trên \(\left[0;2\right]\)

\(f'\left(z\right)=3^z.ln3-4=0\Rightarrow z=log_3\left(\dfrac{4}{ln3}\right)=a\)

\(f\left(0\right)=27\) ; \(f\left(2\right)=27\); \(f\left(a\right)\approx-1,1\)

\(\Rightarrow f\left(z\right)\le27\Rightarrow maxP=27\)

(Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(0;2;2\right)\))

Ồ mà khoan, bài trước bị nhầm lẫn ở chỗ \(3^{2x-x^2}+5^{2y-y^2}\ge3^0+5^0\) mới đúng, ko để ý bị ngược dấu đoạn này

Vậy giải cách khác:

\(0\le x;y;z\le2\Rightarrow x\left(2-x\right)\ge0\Rightarrow2x-x^2\ge0\)

Lại có: \(2x-x^2=1-\left(x-1\right)^2\le1\)

\(\Rightarrow0\le2x-x^2\le1\)

Tương tự ta có: \(0\le2y-y^2\le1\)

Xét hàm: \(f\left(t\right)=3^t-2t\) trên \(\left[0;1\right]\)

\(f'\left(t\right)=3^t.ln3-2=0\Rightarrow t=log_3\left(\dfrac{2}{ln3}\right)=a\)

\(f\left(0\right)=1;\) \(f\left(1\right)=1\) ; \(f\left(a\right)\approx0,73\)

\(\Rightarrow f\left(t\right)\le1\Rightarrow3^t-2t\le1\Rightarrow3^t\le2t+1\)

\(\Rightarrow3^{2x-x^2}\le2\left(2x-x^2\right)+1\)

Hoàn toàn tương tự, ta chứng minh được: 

\(5^t\le4t+1\) với \(t\in\left[0;1\right]\Rightarrow5^{2y-y^2}\le4\left(2y-y^2\right)+1\)

\(3^t\le4t+1\) với \(t\in\left[0;2\right]\Rightarrow3^z\le4z+1\)

\(\Rightarrow P\le2\left(2x-x^2\right)+4\left(2y-y^2\right)+4z+3+2x^2+4y^2=4\left(x+2y+z\right)+3=27\)

Lần này thì ko sai được rồi

sai đề

khong phai sai de dau ban gi oi

\(\frac{16}{3x+3y+2z}=\frac{16}{\left(x+y\right)+\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)

Tương tự:

\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{z+y}\)

\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{z+y}+\frac{1}{y+x}+\frac{1}{x+z}\)

\(\Rightarrow16P\le4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4\cdot6=24\)

\(\Rightarrow P\le\frac{3}{2}\) tại \(x=y=z=\frac{1}{4}\)

We have:

\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)

Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)

Dau '=' xay ra khi \(x=y=z=1\)

Áp dụng bất đẳng thức Cô-si, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3xz+\left(x+y+z\right)\ge3xy+3xz+3\sqrt[3]{xyz}\)\(=3xy+3xz+3\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(xy+xz+1\right)}\)

Tiếp tục áp dụng bất đẳng thức dạng \(u^3+v^3\ge uv\left(u+v\right)\), ta được: \(\frac{1}{3\left(xy+xz+1\right)}=\frac{1}{3\left[x\left(\left(\sqrt[3]{y}\right)^3+\left(\sqrt[3]{z}\right)^3\right)+1\right]}\le\frac{1}{3\left[x\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1\right]}\)\(=\frac{\sqrt[3]{xyz}}{3\left[\sqrt[3]{x^2}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+\sqrt[3]{xyz}\right]}=\frac{\sqrt[3]{yz}}{3\left(\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}\right)}\)

Tương tự rồi cộng lại theo vế, ta được: \(P\le\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z = 1