Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

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 \(x=y=\dfrac{1}{2}\)

\(x^3+y^3+3xy\left(x+y\right)+\dfrac{1}{27}-3xy\left(x+y\right)-xy=0\)

\(\Leftrightarrow\left(x+y\right)^3+\dfrac{1}{27}-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x+y+\dfrac{1}{3}\right)\left[\left(x+y\right)^2-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}\right]-3xy\left(x+y+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x^2+y^2-xy-\dfrac{1}{3}\left(x+y\right)+\dfrac{1}{9}=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(x-\dfrac{1}{3}\right)^2+\left(y-\dfrac{1}{3}\right)^2=0\)

\(\Leftrightarrow x=y=\dfrac{1}{3}\Rightarrow P=...\)

1.

Gọi \(d=ƯC\left(2n^2+3n+1;3n+1\right)\)

\(\Rightarrow2n^2+3n+1-\left(3n+1\right)⋮d\)

\(\Rightarrow2n^2⋮d\Rightarrow2n\left(3n+1\right)-3.2n^2⋮d\)

\(\Rightarrow2n⋮d\Rightarrow2\left(3n+1\right)-3.2n⋮d\Rightarrow2⋮d\Rightarrow\left[{}\begin{matrix}d=1\\d=2\end{matrix}\right.\)

\(d=2\Rightarrow3n+1=2k\Rightarrow n=2m+1\)

\(\Rightarrow n\) lẻ thì A không tối giản

\(\Rightarrow n\) chẵn thì A tối giản

2.

Giả thiết tương đương:

\(xy^2+\dfrac{x^2}{z}+\dfrac{y}{z^2}=3\)

Đặt \(\left(x;y;\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a^2c+b^2a+c^2b=3\)

Ta có: \(9=\left(a^2c+b^2a+c^2b\right)^2\le\left(a^4+b^4+c^4\right)\left(c^2+a^2+b^2\right)\)

\(\Rightarrow9\le\left(a^4+b^4+c^4\right)\sqrt{3\left(a^4+b^4+c^4\right)}\)

\(\Rightarrow3\left(a^4+b^4+c^4\right)^3\ge81\Rightarrow a^4+b^4+c^4\ge3\)

\(\Rightarrow M=\dfrac{1}{a^4+b^4+c^4}\le\dfrac{1}{3}\)

\(M_{max}=\dfrac{1}{3}\) khi \(\left(a;b;c\right)=\left(1;1;1\right)\) hay \(\left(x;y;z\right)=\left(1;1;1\right)\)

Lời giải:

Sửa: $x^2\geq y^2+z^2$
Áp dụng BĐT Cauchy-Schwarz:

$P\geq \frac{y^2+z^2}{x^2}+\frac{7x^2}{2}.\frac{4}{y^2+z^2}+2007$

$=\frac{y^2+z^2}{x^2}+\frac{14x^2}{y^2+z^2}+2007$

$=\frac{y^2+z^2}{x^2}+\frac{x^2}{y^2+z^2}+\frac{13x^2}{y^2+z^2}+2007$

$\geq 2+\frac{13x^2}{y^2+z^2}+2007$ (áp dụng BĐT Cô-si)

$\geq 2+13+2007=2022$ (do $x^2\geq y^2+z^2$)

Vậy $P_{\min}=2022$

Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\)

\(\Leftrightarrow x^2+2\le3x\)

Hoàn toàn tương tự ta có \(y^2+2\le3y\)

Do đó: \(P\ge\dfrac{x+2y}{3x+3y+3}+\dfrac{2x+y}{3x+3y+3}+\dfrac{1}{4\left(x+y-1\right)}\)

\(P\ge\dfrac{x+y}{x+y+1}+\dfrac{1}{4\left(x+y-1\right)}\)

Đặt \(a=x+y-1\Rightarrow1\le a\le3\)

\(\Rightarrow P\ge f\left(a\right)=\dfrac{a+1}{a+2}+\dfrac{1}{4a}\)

\(f'\left(a\right)=\dfrac{3a^2-4a-4}{4a^2\left(a+2\right)^2}=\dfrac{\left(a-2\right)\left(3a+2\right)}{4a^2\left(a+2\right)^2}=0\Rightarrow a=2\)

\(f\left(1\right)=\dfrac{11}{12}\) ; \(f\left(2\right)=\dfrac{7}{8}\) ; \(f\left(3\right)=\dfrac{53}{60}\)

\(\Rightarrow f\left(a\right)\ge\dfrac{7}{8}\Rightarrow P_{min}=\dfrac{7}{8}\) khi \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)

\(4=2^x+2^y\ge2\sqrt{2^{x+y}}\Rightarrow2^{x+y}\le4\Rightarrow x+y\le2\)

\(\Rightarrow xy\le1\)

\(P=4x^2y^2+2x^3+2y^3+10xy\)

\(P=4x^2y^2+10xy+2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)

\(P\le4x^2y^2+10xy+4\left(4-3xy\right)=4x^2y^2-2xy+16\)

Đặt \(xy=t\Rightarrow0< t\le1\)

Xét hàm \(f\left(t\right)=4t^2-2t+16\) trên \((0;1]\)

\(\Rightarrow...\)