Lời giải:
\(A=\frac{x^2}{\sqrt{x^4+8xy^3}}+\frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\)

Xét:

\(x^4+8xy^3-(x^2+2y^2)^2=8xy^3-4y^4-4x^2y^2\)

\(=-4y^2(x^2-2xy+y^2)=-4y^2(x-y)^2\leq 0\)

\(\Rightarrow x^4+8xy^3\leq (x^2+2y^2)^2\)

\(\Rightarrow \frac{x^2}{\sqrt{x^4+8xy^3}}\geq \frac{x^2}{x^2+2y^2}(*)\)

Mặt khác:
\(y^4+y(x+y)^3-(x^2+2y^2)^2=x^3y+3xy^3-2y^4-x^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^4-x^2y^2\)

\(=x^3(y-x)+3y^3(x-y)+y^2(y-x)(y+x)\)

\(=(y-x)(x^3-2y^3+xy^2)\)

\(=(y-x)[(x-y)(x^2+xy+y^2)+y^2(x-y)]\)

\(=-(x-y)^2(x^2+xy+2y^2)\leq 0\)

\(\Rightarrow y^4+y(x+y)^3\leq (x^2+2y^2)^2\Rightarrow \frac{2y^2}{\sqrt{y^4+y(x+y)^3}}\geq \frac{2y^2}{x^2+2y^2}(**)\)

Từ $(*); (**)\Rightarrow A\geq 1$

\(T=\sqrt{\dfrac{x^3}{x^3+8y^3}}+\sqrt{\dfrac{4y^3}{y^3+\left(x+y\right)^3}}\)

\(=\dfrac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\dfrac{2y^2}{\sqrt{y\left(y^3+\left(x+y\right)^3\right)}}\)

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

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

\(\ge\dfrac{2x^2+4y^2}{2x^2+4y^2}=1\)

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 \(x=y=z=\dfrac{1}{2}\)

Lời giải:

Đặt $\frac{y}{x}=a(a>0)$ thì:

\(P=\sqrt{\frac{1}{1+(\frac{2y}{x})^3}}+\sqrt{\frac{4}{1+(1+\frac{x}{y})^3}}=\sqrt{\frac{1}{1+8a^3}}+\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\)

Áp dụng BĐT AM-GM dạng $xy\leq \left(\frac{x+y}{2}\right)^2$ ta có:

\(1+8a^3=1+(2a)^3=(1+2a)(1-2a+4a^2)\leq \left(\frac{1+2a+1-2a+4a^2}{2}\right)^2=(2a^2+1)^2\)

\(\Rightarrow \sqrt{\frac{1}{8a^3+1}}\geq \frac{1}{2a^2+1}(1)\)

\(1+(1+\frac{1}{a})^3=(2+\frac{1}{a})[1-(1+\frac{1}{a})+(1+\frac{1}{a})^2]\leq (\frac{3a^2+2a+1}{2a^2})^2\)

\(\Rightarrow \sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{4a^2}{3a^2+2a+1}\)

Mà: \(\frac{4a^2}{3a^2+2a+1}\geq \frac{4a^2}{3a^2+a^2+1+1}=\frac{2a^2}{2a^2+1}\) nên \(\sqrt{\frac{4}{1+(1+\frac{1}{a})^3}}\geq \frac{2a^2}{2a^2+1}(2)\)

Từ $(1);(2)\Rightarrow P\geq \frac{1}{2a^2+1}+\frac{2a^2}{2a^2+1}=1$

Vậy $P_{\min}=1$ khi $a=1\Leftrightarrow x=y$