\(a^2+b^2>=2ab\)

\(b^2+c^2>=2bc\)

\(a^2+c^2>=2ac\)

=> \(2\left(a^2+b^2+c^2\right)>=2\left(ab+bc+ac\right)\)DẤU '=' xảy ra khi a=b=c

\(a< \sqrt{ab}\)

\(\Leftrightarrow a^2< ab\)

\(\Leftrightarrow a^2-ab< 0\)

\(\Leftrightarrow a\left(a-b\right)< 0\) (đúng) (1)

\(\sqrt{ab}< \dfrac{a+b}{2}\) (áp dụng BĐT AM-GM). (2)

\(\dfrac{a+b}{2}< b\)

\(\Leftrightarrow\dfrac{a}{2}-\dfrac{b}{2}< 0\)

\(\Leftrightarrow\dfrac{a-b}{2}< 0\) (đúng) (3)

-Từ (1), (2), (3) ta suy ra đpcm.

a/b+b/a-ab

=a/b+b/a-(a-b)

=a/b+b/a-a+b

=a/b-a+b/a+b

=(a-ab)/b+(b+ab/a)

=(a-a+b)/b-((b+a-b)a

=1+1

=2

vì a,b khác  0 => a.b khác 0

ta có: a/b + b/a - ab

=(a^2+b^2-a^2b^2)/ab

=[(a-b)^2+2ab-a^2b^2]/ab

=(a^2b^2+2ab-a^2b^2)/ab=2ab/ab=2 (do a-b=ab)

1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:

\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).

Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).

2.

\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)

Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)

\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )

\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)

\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)

Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)

3. Chia 2 vế giả thiết cho \(x^2y^2\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)