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\(1\le5\left(x+y\right)+4\left(x+y\right)^2+9xy\le5\left(x+y\right)+4\left(x+y\right)^2+\frac{9}{4}\left(x+y\right)^2\)
\(\Leftrightarrow25\left(x+y\right)^2+20\left(x+y\right)-4\ge0\)
\(\Rightarrow x+y\ge\frac{2\sqrt{2}-2}{5}\)
\(P=17\left(x+y\right)^2-18xy\ge17\left(x+y\right)^2-\frac{9}{2}\left(x+y\right)^2=\frac{25}{2}\left(x+y\right)^2\ge\frac{25}{2}\left(\frac{2\sqrt{2}-2}{5}\right)^2=6-4\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{\sqrt{2}-1}{5}\)
\(P=x^2+y^2+z^2\ge\dfrac{1}{3}\left(x+y+z\right)^3=\dfrac{64}{3}\)
\(P_{min}=\dfrac{64}{3}\) khi \(x=y=z=\dfrac{4}{3}\)
Đặt \(\left(x;y;z\right)=\left(a+1;b+1;c+1\right)\Rightarrow\left\{{}\begin{matrix}a+b+c=1\\a;b;c\ge0\end{matrix}\right.\)
\(\Rightarrow0\le a;b;c\le1\) \(\Rightarrow\left\{{}\begin{matrix}a^2\le a\\b^2\le b\\c^2\le c\end{matrix}\right.\) \(\Rightarrow a^2+b^2+c^2\le a+b+c=1\)
\(P=\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2\)
\(P=a^2+b^2+c^2+2\left(a+b+c\right)+3=a^2+b^2+c^2+5\le1+5=6\)
\(P_{max}=6\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và hoán vị hay \(\left(x;y;z\right)=\left(1;1;2\right)\) và hoán vị
\(A=x^2+3xy+4y^2=\frac{7}{16}x^2+\frac{9}{16}x^2+3xy+4y^2=\frac{7}{16}x^2+\left(\frac{3}{4}x+2y\right)^2\)
\(\ge\frac{7}{16}.1^2+0^2=\frac{7}{16}\)
Dấu \(=\)khi \(\hept{\begin{cases}x=1\\\frac{3}{4}x+2y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-\frac{3}{8}\end{cases}}\).
Ta co A = 2(x+y)+\(\frac{2}{x+y}\)\(\ge2\sqrt{2\left(x+y\right).\frac{2}{x+y}}\)=4 khi x=y =\(\frac{1}{2}\)
ta dễ chứng minh được \(x+y\ge\frac{2\sqrt{2}}{5}-\frac{2}{5}\)\(\Rightarrow\)\(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}>0\)
\(P=\frac{5\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\left(\frac{5}{2}\left(x+y-\left(\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)\right)\left(\frac{5}{2}\left(x+y\right)+\sqrt{2}+1\right)-\frac{9}{4}\left(x-y\right)^2\right)}{\frac{5}{2}\left(x+y\right)+\sqrt{2}+1}\)
\(+\left(\frac{\frac{45}{2}\left(x+y+\frac{2\sqrt{2}}{5}-\frac{2}{5}\right)}{5\left(x+y\right)+\sqrt{2}+1}+\frac{9}{2}\right)\left(x-y\right)^2+6-4\sqrt{2}\ge6-4\sqrt{2}\)
Dấu "=" xảy ra khi \(x=y=\frac{\sqrt{2}-1}{5}\)
Ta chứng minh: \(P\ge6-4\sqrt{2}+\left(2-\sqrt{2}\right)\left(4x^2+4y^2+17xy+5x+5y-11\right)\)
Hay là:
\(\frac{\left(9+4\sqrt{2}\right)\left(98x-298y-130+225\sqrt{2}y+85\sqrt{2}\right)^2}{9604}+\frac{18\left(2\sqrt{2}-1\right)\left(-5y-1+\sqrt{2}\right)^2}{36+16\sqrt{2}}\ge0\)
Việc còn lại là của mọi người.