Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{x^4}{a}+\frac{y^4}{b}\right)(a+b)\geq (x^2+y^2)^2=1\)

\(\Leftrightarrow \frac{x^4}{a}+\frac{y^4}{b}\geq \frac{1}{a+b}\)

Dấu bằng xảy ra khi \(\frac{x^2}{a}=\frac{y^2}{b}\). Do đó \(\frac{x^2}{a}=\frac{y^2}{b}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow \frac{x^{2006}}{a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{(a+b)^{1003}}\)

\(\Rightarrow \frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{y^{1003}}=\frac{2}{(a+b)^{1003}}\)

Do đó ta có đpcm.

 \(\text{Đặt }x^2=m\ge0;y^2=n\ge0\Rightarrow m+n=1\)

\(\text{Ta có: }\frac{m^2}{a}+\frac{n^2}{b}=\frac{\left(m+n\right)^2}{a+b}\Leftrightarrow\left(a+b\right)\left(\frac{m^2}{a}+\frac{n^2}{b}\right)=\left(m+n\right)^2\left(\text{BĐT Bunhiacopki}\right)\)\(\Leftrightarrow m^2+n^2+\frac{b}{a}m^2+\frac{a}{b}n^2=m^2+n^2+2mn\)

\(\Leftrightarrow\frac{b}{a}m^2+\frac{a}{b}n^2-2mn=0\left(1\right)\)

\(\text{+Nếu }\frac{a}{b}< 0\text{ thì (1)}\Leftrightarrow-\left(\sqrt{-\frac{b}{a}m}\right)^2-2mn-\left(\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}\right)^2=0\)

\(\Leftrightarrow\sqrt{-\frac{b}{a}m}+\sqrt{-\frac{a}{b}n}=0\Leftrightarrow m=n=0\left(\text{loại}\right)\)

\(\text{Xét }\frac{a}{b}>0;\left(1\right)\Leftrightarrow\left(\sqrt{\frac{b}{a}m}\right)^2-2mn+\left(\sqrt{\frac{a}{b}n}\right)^2=0\)

\(\Leftrightarrow\left(\sqrt{-\frac{b}{a}m}-\sqrt{-\frac{a}{b}n}\right)^2=0\Leftrightarrow\sqrt{\frac{b}{a}m}=\sqrt{\frac{a}{b}n}\)

\(\Leftrightarrow bm=an\Leftrightarrow bx^2=ay^2\left(a,b>0\right)\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\left(\frac{x^2}{a}\right)^{1003}+\left(\frac{y^2}{b}\right)^{1003}=\frac{1}{\left(a+b\right)^{1003}}+\frac{1}{\left(a+b\right)^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\left(đpcm\right)\)

5a.

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+....+\dfrac{1}{19.21}\\ =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{21}\right)\\ =\dfrac{1}{2}.\dfrac{20}{21}=\dfrac{10}{21}\)

b.

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\\ =\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+....+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\\ =\dfrac{1}{2}\left(1-\dfrac{1}{2n+1}\right)< \dfrac{1}{2}.1=\dfrac{1}{2}\)

1/ Ta có: \(\frac{x^4}{1a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\)

\(\Leftrightarrow1bx^4\left(a+b\right)+ay^4\left(a+b\right)=ab\left(x^4+2x^2y^2+y^4\right)\)

 \(\Leftrightarrow\left(ay^2-bx^2\right)^2=0\)

\(\Rightarrow\frac{x^2}{1a}=\frac{y^2}{b}=\frac{\left(x^2+y^2\right)}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\frac{x^{2006}}{1a^{1003}}=\frac{y^{2006}}{b^{1003}}=\frac{1}{\left(a+b\right)^{1003}}\)

 \(\Rightarrow\frac{x^{2006}}{a^{1003}}+\frac{y^{2006}}{b^{1003}}=\frac{2}{\left(a+b\right)^{1003}}\)

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 \(x=y=\dfrac{1}{2}\)