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 \(x=y=\dfrac{1}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel có:

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=6\)

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

áp dụng BDT AM-GM

\(=>x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\sqrt{xy}\le\dfrac{1}{2}=>xy\le\dfrac{1}{4}\)

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

\(\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{1}{4}}=\dfrac{4}{\left(x+y\right)^2}+2=4+2=6\)

dấu"=" xảy ra \(< =>x=y=\dfrac{1}{2}\)

h

\(2=3\sqrt{xy}+2\sqrt{xz}\le\dfrac{3}{2}\left(x+y\right)+x+z\)

\(\Rightarrow5x+3y+2z\ge4\)

\(A=5\left(\dfrac{xy}{z}+\dfrac{xz}{y}\right)+3\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+2\left(\dfrac{xz}{y}+\dfrac{yz}{x}\right)\)

\(A\ge5.2x+3.2y+2.2z=2\left(5x+3y+2z\right)\ge8\)

\(A_{min}=8\) khi \(x=y=z=\dfrac{2}{5}\)

Với a,b,c dưog thì \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}>=\dfrac{\left(x+y+z\right)^2}{a+b+c}\)

\(P>=\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+\sqrt{1+x^3}+\sqrt{1+y^3}+\sqrt{1+z^3}}\)

\(\sqrt{1+x^3}=\sqrt{\left(1+x\right)\left(1-x+x^2\right)}< =\dfrac{2+x^2}{2}\)

Dấu = xảy ra khi x=2

=>\(P>=\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2+6}=\dfrac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2+6}\)

Đặt t=(x+y+z)^2(t>=36)

=>P>=2t/t-6

Xét hàm số \(f\left(t\right)=\dfrac{t}{t+6}\left(t>=36\right)\)

\(f'\left(t\right)=\dfrac{6}{\left(t+6\right)^2}>=0,\forall t>=36\)

=>f(t) đồng biến

=>f(t)>=f(36)=6/7

=>P>=12/7

Dấu = xảy ra khi x=y=z=2

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

Áp dụng BĐT AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)

Ta chứng minh BĐT sau:

Ta có: \(x\left(3-4x^2\right)=-4x^3+3x-1+1=1-\left(x+1\right)\left(2x-1\right)^2\le1\)

\(\Rightarrow\dfrac{4x^2}{x\left(3-4x^2\right)}\ge\dfrac{4x^2}{1}=4x^2\)

Tương tự và cộng lại:

\(Q\ge4\left(x^2+y^2+z^2\right)\ge4\left(xy+yz+zx\right)=3\)

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