2.
Áp dụng BĐT \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow VT=\sqrt{2x+1}+\sqrt{2y+1}+\sqrt{2z+1}\le\sqrt{3\left(2x+1+2y+1+2z+1\right)}\)
\(\Rightarrow VT\le\sqrt{3\left[2\left(x+y+z\right)+3\right]}=\sqrt{15}< \sqrt{16}=4\) (đpcm)
3.
\(VT=a^4+b^4+c^4\ge\frac{1}{3}\left(a^2+b^2+c^2\right)^2\ge\frac{1}{3}\left[3\left(ab+bc+ca\right)\right]^2=27\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)