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

\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)

Dấu "=" xảy ra khi \(x=y\)

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

\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)

\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)

\(B_{min}=1\) khi \(x=y\)

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 \(x=y=\dfrac{1}{2}\)

\(A=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}+\dfrac{3\left(x^2+y^2\right)}{4xy}\)

\(A\ge2\sqrt{\dfrac{\left(x^2+y^2\right)xy}{4xy\left(x^2+y^2\right)}}+\dfrac{3.2xy}{4xy}=\dfrac{5}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(C=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{6xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{6xy}{\left(x+y\right)^2}-4\)

\(C=\dfrac{3\left(x+y\right)^2}{8xy}+\dfrac{6xy}{\left(x+y\right)^2}+\dfrac{5\left(x+y\right)^2}{8xy}-4\)

\(C\ge2\sqrt{\dfrac{18xy\left(x+y\right)^2}{8xy\left(x+y\right)^2}}+\dfrac{5.4xy}{8xy}-4=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y\)

\(1,A=5^{n+2}+26\cdot5^n+8^{2n+1}\\ A=5^n\cdot25+26\cdot5^n+8\cdot8^{2n+1}\\ A=51\cdot5^n+8\cdot64^n\)

Ta có \(64:59R5\Rightarrow64^n:59R5\)

Vì vậy \(51\cdot5^n+8\cdot64^n:59R=5^n\cdot51+8\cdot5^n=5^n\left(51+8\right)=5^n\cdot59⋮59\)

Vậy \(A⋮59\)

(\(R\) là dư)

\(2,\\ a,2x\ge0;\left(x+2\right)^2\ge0,\forall x\\ \Leftrightarrow P=\dfrac{\left(x+2\right)^2}{2x}\ge0\\ P_{min}=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\)

\(\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{1}{4xy}.4xy}+\dfrac{5}{4.\dfrac{\left(x+y\right)^2}{4}}\)

\(\ge\dfrac{4}{1^2}+2+\dfrac{5}{1^2}\) (do \(x+y\le1\))

\(=11\)

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

Vậy GTNN của A là 11.

Tương tự, ta được:

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

và \(\left(2-z\right)\left(2-x\right)>=\left(\dfrac{y+1}{2}\right)^2\)

=>8(2-x)(2-y)(2-z)>=(x+1)(y+1)(z+1)

(x+yz)(y+zx)<=(x+y+yz+xz)^2/4=(x+y)^2*(z+1)^2/4<=(x^2+y^2)(z+1)^2/4

Tương tự, ta cũng co:

\(\left(y+xz\right)\left(z+y\right)< =\dfrac{\left(y^2+z^2\right)\left(x+1\right)^2}{2}\)

và \(\left(z+xy\right)\left(x+yz\right)< =\dfrac{\left(z^2+x^2\right)\left(y+1\right)^2}{2}\)

Do đó, ta được:

\(\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)< =\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

=>ĐPCM

\(M=3\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)+\dfrac{1}{2xy}\ge\dfrac{12}{2xy+x^2+y^2}+\dfrac{2}{\left(x+y\right)^2}=\dfrac{14}{\left(x+y\right)^2}=14\)

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

Áp dụng bđt đã cho ta có \(M=4\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)-\dfrac{1}{x^2+y^2}\ge\dfrac{16}{2xy+x^2+y^2}-\dfrac{2}{\left(x+y\right)^2}=\dfrac{16}{\left(x+y\right)^2}-\dfrac{2}{\left(x+y\right)^2}=14\).

Đẳng thức xảy ra khi và chỉ khi \(x=y=\dfrac{1}{2}\)

Bài 3:

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

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

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)

\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)

\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)

-Tham khảo:

-Tham khảo:

\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)

\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)

\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)