\(A=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\left(\dfrac{1}{2xy}+8xy\right)+\dfrac{3}{xy}\)

\(A\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{8xy}{2xy}}+\dfrac{3}{\dfrac{1}{4}\left(x+y\right)^2}\ge20\)

\(A_{min}=20\) khi \(x=y=\dfrac{1}{2}\)

cau 2 chung minh cai gi vay ban

\(P=x^2+y^2=\left(x+y\right)^2-2xy=\left(15-xy\right)^2-2xy=\left(xy\right)^2-32xy+225=p^2-32p+225.\)

s+p = 15 ; s² -4p>/ 0 => p</ 3

P min = 138 khi  p = 3 ; s = 12 

bạn gải thích rõ bước cuối được không bạn từ bước s+p=15;s²-4p>/0

Biến đổi từ giả thiết

\(x^3+y^3+6xy\le8\)

\(\Leftrightarrow...\Leftrightarrow\left(x+y-2\right)\left(x^2-xy+y^2+2x+2y+4\right)\le0\)

\(\Leftrightarrow x+y-2\le0\)

(Do \(x^2-xy+y^2+2x+2y+4=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4}+2x+2y+4>0\forall x;y>0\))

\(\Leftrightarrow x+y\le2\)

Và áp dụng các bđt \(\frac{1}{2ab}\ge\frac{2}{\left(a+b\right)^2}\)

                                 \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\left(a;b>0\right)\)

Khi đó \(P=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(\frac{1}{ab}+ab\right)+\frac{3}{2ab}\)

               \(\ge\frac{4}{a^2+b^2+2ab}+2+\frac{6}{\left(a+b\right)^2}\)

                 \(=\frac{4}{\left(a+b\right)^2}+2+\frac{6}{\left(a+b\right)^2}\ge\frac{9}{2}\)

Dấu "=" <=> a= b = 1

\(x\ge xy+1\Rightarrow1\ge y+\dfrac{1}{x}\ge2\sqrt{\dfrac{y}{x}}\Rightarrow\dfrac{y}{x}\le\dfrac{1}{4}\)

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

Đặt \(\dfrac{y}{x}=t\le\dfrac{1}{4}\) 

\(Q^2=\dfrac{t^2+2t+1}{t^2-t+3}=\dfrac{t^2+2t+1}{t^2-t+3}-\dfrac{5}{9}+\dfrac{5}{9}\)

\(Q^2=\dfrac{\left(4t-1\right)\left(t+6\right)}{9\left(t^2-t+3\right)}+\dfrac{5}{9}\le\dfrac{5}{9}\)

\(\Rightarrow Q_{max}=\dfrac{\sqrt{5}}{3}\) khi \(t=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(2;\dfrac{1}{2}\right)\)