Question

[NGUYỄN DOÃN ANH THÁI]

Cho x,y,z \(\ge\)0 thoả mãn x+y+z \(\le\)3. Tìm Max của

\(\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+3\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

Answer 1

[alibaba nguyễn]

Ta có \(\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)

\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\)

\(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)

\(\Rightarrow\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{2}\left(x+y+z+3\right)\le6\sqrt{2}\)

Ta lại có \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}\le3\)

Theo đề bài ta có

\(\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+3\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)

\(\le6\sqrt{2}+\left(3-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le3\sqrt{2}+9\)

Dấu = xảy ra khi x = y = z = 1