a) câu a này chắc thiếu đề hay sai gì gì đó rồi
chứ x2+y2+1\(\ge1\forall x;y\) mà
b)Áp dụng bđt Bunhiacopxki:
\(\left(a+b+c\right)^2\le\left(a^2+b^2+c^2\right)\left(1+1+1\right)=3\left(a^2+b^2+c^2\right)\)
<=>\(1^2\le3\left(a^2+b^2+c^2\right)\Leftrightarrow a^2+b^2+c^2\ge\dfrac{1}{3}\)
Tiếp tục áp dụng bđt Bunhiacopxki:
\(\left(a^2+b^2+c^2\right)^2\le\left(a^4+b^4+c^4\right)\left(1+1+1\right)=3\left(a^4+b^4+c^4\right)\)
<=>\(3\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\ge\left(\dfrac{1}{3}\right)^2\)
<=>\(3\left(a^4+b^4+c^4\right)\ge\dfrac{1}{9}\Leftrightarrow a^4+b^4+c^4\ge\dfrac{1}{27}\)
Dấu "=" xảy ra khi a=b=c=1/3
đề câu a như bạn nói: \(x^2+y^2+1\ge xy+x+y\) giải như này:
\(x^2+y^2+1\ge xy+x+y\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)
<=>\(2x^2+2y^2+2\ge2xy+2x+2y\)
\(\Leftrightarrow2x^2+2y^2+2-2xy-2x-2y\ge0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\) đúng do \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\)
Dấu "=" xảy ra khi \(\left(x-y\right)^2=\left(x-1\right)^2=\left(y-1\right)^2=0\)
<=>\(x-y=x-1=y-1=0\Leftrightarrow x=y=1\)
