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\(9=3a^2+2b^2+2bc+2c^2=\left(a+b+c\right)^2+2a^2+b^2+c^2-2a\left(b+c\right)\)
\(\Rightarrow9\ge\left(a+b+c\right)^2+2a^2+\dfrac{1}{2}\left(b+c\right)^2-2a\left(b+c\right)\)
\(\Rightarrow9\ge\left(a+b+c\right)^2+\dfrac{1}{2}\left(2a-b-c\right)^2\ge\left(a+b+c\right)^2\)
\(\Rightarrow-3\le a+b+c\le3\)
\(T_{max}=3\) khi \(a=b=c=1\)
\(T_{min}=-3\) khi \(a=b=c=-1\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ; \(\forall a;b;c\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow ab+bc+ca\le1\)
\(\Rightarrow P_{max}=1\) khi \(a=b=c\)
Lại có:
\(\left(a+b+c\right)^2\ge0\) ; \(\forall a;b;c\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow ab+bc+ca\ge-\dfrac{a^2+b^2+c^2}{2}=-\dfrac{1}{2}\)
\(P_{min}=-\dfrac{1}{2}\) khi \(a+b+c=0\)
Cho hai số a,b thỏa mãn a^2+b^2=1. Tìm giá trị lớn nhất và giá trị nhỏ nhất của biểu thức: A=a^6+b^6
\(A=a^6+b^6=\left(a^2\right)^3+\left(b^2\right)^3\)
\(=\left(a^2+b^2\right)\left(a^4+b^4-a^2b^2\right)\)
\(=1.\left[\left(a^4+b^4+2a^2b^2\right)-3a^2b^2\right]\)
\(=\left(a^2+b^2\right)^2-3a^2b^2\)
\(=1^2-3a^2b^2\)
\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2-2ab\ge0\Rightarrow\frac{a^2+b^2}{2}\ge ab\)
\(\Rightarrow ab\le1:2=0,5\Rightarrow3a^2b^2\le\frac{3}{4}\)
\(\Rightarrow A=1^2-3a^2b^2\ge1-\frac{3}{4}=\frac{1}{4}\)
\(\Rightarrow MinA=\frac{1}{4}\Leftrightarrow a=b=\frac{1}{2}\)
Vậy ...
Từ giả thiết:
\(a^2=2\left(b^2+c^2\right)\ge\left(b+c\right)^2\Rightarrow\left(\dfrac{a}{b+c}\right)^2\ge1\Rightarrow\dfrac{a}{b+c}\ge1\)
\(P=\dfrac{a}{b+c}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+2bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+\dfrac{1}{2}\left(b+c\right)^2}\)
\(P\ge\dfrac{a}{b+c}+\dfrac{1}{\dfrac{a}{b+c}+\dfrac{1}{2}}\)
Đặt \(\dfrac{a}{b+c}=x\ge1\)
\(\Rightarrow P\ge x+\dfrac{1}{x+\dfrac{1}{2}}=\dfrac{4}{9}\left(x+\dfrac{1}{2}\right)+\dfrac{1}{x+\dfrac{1}{2}}+\dfrac{5}{9}x-\dfrac{2}{9}\)
\(P\ge2\sqrt{\dfrac{4}{9}\left(x+\dfrac{1}{2}\right).\dfrac{1}{\left(x+\dfrac{1}{2}\right)}}+\dfrac{5}{9}.1-\dfrac{2}{9}=\dfrac{5}{3}\)
\(P_{min}=\dfrac{5}{3}\) khi \(x=1\) hay \(a=2b=2c\)
\(a+b\ge a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\)
\(\Rightarrow2\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le1\)
Xét \(Q=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a\left(b+1\right)+b\left(a+1\right)}{\left(a+1\right)\left(b+1\right)}=\dfrac{a+b+2ab}{\left(a+1\right)\left(b+1\right)}\)
\(Q=\dfrac{a+b+ab+ab}{\left(a+1\right)\left(b+1\right)}\le\dfrac{a+b+ab+1}{\left(a+1\right)\left(b+1\right)}=\dfrac{\left(a+1\right)\left(b+1\right)}{\left(a+1\right)\left(b+1\right)}=1\)
\(\Rightarrow P\le2020+1^{2021}=2021\)
Dấu "=" xảy ra khi \(a=b=1\)
\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Leftrightarrow4\ge\left(a+b\right)^2\Leftrightarrow-2\le a+b\le2\)
\(4ab\le2\left(a^2+b^2\right)\Leftrightarrow4ab\le4\Leftrightarrow ab\le1\)
\(A=\left(3-a\right)\left(3-b\right)=9-3a-3b+ab=9-3\left(a+b\right)+ab\le9+3.2+1=16\)
\(A_{max}=16\Leftrightarrow a=b=-1\)
\(\left(a+b\right)^2\ge2\left(a^2+b^2\right)=4\Rightarrow-2\le a+b\le2\)
\(P=9-3\left(a+b\right)+ab=9-3\left(a+b\right)+\dfrac{\left(a+b\right)^2-\left(a^2+b^2\right)}{2}\)
\(P=\dfrac{1}{2}\left(a+b\right)^2-3\left(a+b\right)+8\)
Đặt \(a+b=x\Rightarrow-2\le x\le2\)
\(P=\dfrac{1}{2}x^2-3x+8=\dfrac{1}{2}\left(x-2\right)\left(x-4\right)+4\)
Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x-2\le0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x-2\right)\left(x-4\right)\ge0\)
\(\Rightarrow P\ge4\Rightarrow P_{min}=4\) khi \(x=2\Leftrightarrow a=b=1\)
\(P=\dfrac{1}{2}\left(x+2\right)\left(x-8\right)+16\)
Do \(-2\le x\le2\Rightarrow\left\{{}\begin{matrix}x+2\ge0\\x-8< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{2}\left(x+2\right)\left(x-8\right)\le0\)
\(\Rightarrow P\le16\Rightarrow P_{max}=16\) khi \(x=-2\Leftrightarrow a=b=-1\)