ta có : a = 5 - b \(\Rightarrow a+b=5\)

theo bđt cô si ta có : \(\frac{a+b}{2}\ge\sqrt{ab}\)

                  \(\Leftrightarrow\left(\frac{a+b}{2}\right)^2\ge ab\)

                  \(\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge ab\)

                  \(\Leftrightarrow\frac{25}{4}\ge ab\) ab đạt GTNN khi ab = \(\frac{25}{4}\)

                 ta có   \(\frac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow a+b\ge2\sqrt{ab}\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}}\)

                  \(P\ge\frac{2}{\sqrt{ab}}\Leftrightarrow P\ge\frac{2}{\sqrt{\frac{25}{4}}}=\frac{4}{5}\)

dấu " = " xảy ra khi P = 4/5

mik làm lụi >_< 

Bạn xem lại xem viết đề đã đúng chưa vậy?

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)

Ta dễ có:

\(2+4ab=\left(a+b\right)^2+a+b\ge4ab+a+b\Rightarrow a+b\le2\)

\(P=\frac{a^2-2a+2}{b+1}+\frac{b^2-2b+2}{a+1}\)

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

\(\ge\frac{\left(a+b-2\right)^2}{a+b+2}+\frac{4}{a+b+2}\ge\frac{\left(a+b-2\right)^2}{a+b+2}+1\ge1\)

Đẳng thức xảy ra tại \(a=b=1\)

hmm check hộ mình nhá

\(P=\frac{1}{a^2+b^2+1}+\frac{1}{2ab}\)

\(P=\frac{1}{a^2+b^2+1}+\frac{\frac{1}{9}}{2ab}+\frac{4}{9ab}\)

\(\ge\frac{\left(1+\frac{1}{3}\right)^2}{a^2+b^2+1+2ab}+\frac{4}{9ab}\)

\(\ge\frac{\left(1+\frac{3}{4}\right)^2}{\left(a+b\right)^2+1}+\frac{16}{9\left(a+b\right)^2}\)

\(\ge\frac{\left(1+\frac{1}{3}\right)^2}{1+1}+\frac{16}{9}=\frac{8}{3}\)

Dấu = xảy ra khi \(a=b=\frac{1}{2}\)