hình như là \(x\in R\)

\(a,\frac{a+b}{2}\ge\sqrt{ab}\Rightarrow a+b\ge2\sqrt{ab}\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\)

\(\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow a^2-2ab+b^2\ge0\)

\(\Rightarrow\left(a-b\right)^2\ge0\)( luôn đúng ) 

\(\Rightarrow\frac{a+b}{2}\ge\sqrt{ab}\)

Lời giải :

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)

\(=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{16}-\sqrt{15}}{\left(\sqrt{15}+\sqrt{16}\right)\left(\sqrt{16}-\sqrt{15}\right)}\)

\(=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{16}-\sqrt{15}}{16-15}\)

\(=\frac{\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{16}-\sqrt{15}}{1}\)

\(=\sqrt{16}-\sqrt{1}\)

\(=4-1=3\)

\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{15}+\sqrt{16}}\)

\(=-\left(\sqrt{1}-\sqrt{2}\right)-\left(\sqrt{2}-\sqrt{3}\right)-...-\left(\sqrt{15}-\sqrt{16}\right)\)

\(=-\left(\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+...+\sqrt{15}-\sqrt{16}\right)\)

\(=-\left(\sqrt{1}-\sqrt{16}\right)=-\left(1-4\right)=3\)

Giá trị nhỏ nhất nha các bạn !

\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

= \(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}.1}+1}}\)

=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\left|\sqrt{3}+1\right|}}\)

=\(\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)

=\(\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)

=\(\sqrt{6+2.\left(\sqrt{2}.\sqrt{2-\sqrt{3}}\right)}\)

=\(\sqrt{6+2.\left(\sqrt{4-2\sqrt{3}}\right)}\)

=\(\sqrt{6+2.\sqrt{\left(\sqrt{3}-1\right)^2}}\)

=\(\sqrt{6+2.\left|\sqrt{3}-1\right|}\)

=\(\sqrt{6+2\sqrt{3}-2}\)

=\(\sqrt{4+2\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{3}+1\right)^2}\)

=\(\left|\sqrt{3}+1\right|\)

=\(\sqrt{3}+1\)