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Không mất tính tổng quát, giả sử \(a\ge b\ge c\)
\(\Rightarrow P\le\dfrac{a}{b+c+1}+\dfrac{b}{b+c+1}+\dfrac{c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
\(\Rightarrow P\le\dfrac{a+b+c}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)=\dfrac{a-1}{b+c+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)+1\)
\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{b+c+1}\right]+1\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{bc+b+c+1}\right]+1\)
\(\Rightarrow P\le\left(1-a\right)\left[\left(1-b\right)\left(1-c\right)-\dfrac{1}{\left(1+b\right)\left(1+c\right)}\right]+1\)
\(\Rightarrow P\le\left(1-a\right)\left(\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right)+1\)
Do \(a;b;c\le1\Rightarrow\left\{{}\begin{matrix}1-a\ge0\\\left(1-b^2\right)\left(1-c^2\right)\le1\\\end{matrix}\right.\) \(\Rightarrow\left(1-a\right)\left[\dfrac{\left(1-b^2\right)\left(1-c^2\right)-1}{\left(1+b\right)\left(1+c\right)}\right]\le0\)
\(\Rightarrow P\le1\)
\(P_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;0\right);\left(1;1;1\right);\left(0;1;1\right);\left(0;0;1\right)\) và các hoán vị
a>1; b>1/2; c>1/3 và \(\frac{1}{a} + \frac{2}{2b + 1} + \frac{3}{3c + 2} \ge 2 \quad (1)\)
Đặt x=a
=>x>1
=>x-1>0
Đặt y=2b+1
=>y>2*1/2+1=1+1=2
=>y-2>0
Đặt z=3c+2
=>z>3
=>z-3>0
P=(a-1)(2b-1)(3c-1)
=(x-1)(y-2)(z-3)
(1) =>\(\frac{1}{x} + \frac{2}{y} + \frac{3}{z} \ge 2 \quad (2)\)
=>\(\frac{1}{x} \ge 2 - \frac{2}{y} - \frac{3}{z} = \left(1 - \frac{2}{y}\right) + \left(1 - \frac{3}{z}\right) = \frac{y - 2}{y} + \frac{z - 3}{z}\)
=>\(\frac{1}{x} \ge \frac{y - 2}{y} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \quad (3)\)
Chứng minh tương tự, ta sẽ có:
\(\frac{2}{y} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{3}{z}\right) = \frac{x - 1}{x} + \frac{z - 3}{z} \ge 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \quad (4)\)
\(\frac{3}{z} \ge \left(1 - \frac{1}{x}\right) + \left(1 - \frac{2}{y}\right) = \frac{x - 1}{x} + \frac{y - 2}{y} \ge 2\sqrt{\frac{(x - 1)(y - 2)}{xy}} \quad (5)\)
Từ (3),(4),(5) suy ra \(\frac{1}{x} \cdot \frac{2}{y} \cdot \frac{3}{z} \ge 2\sqrt{\frac{(y - 2)(z - 3)}{yz}} \cdot 2\sqrt{\frac{(x - 1)(z - 3)}{xz}} \cdot 2\sqrt{\frac{(x - 1)(y - 2)}{xy}}\)
=>\(\frac{6}{xyz}\ge8\cdot\frac{(x - 1)(y - 2)(z - 3)}{xyz}\)
=>\(6\ge8(x-1)(y-2)(z-3)\)
=>\((x-1)(y-2)(z-3)\le\frac{6}{8}=\frac{3}{4}\)
=>P<=3/4
Dấu '=' xảy ra khi \(\begin{cases} \dfrac{x - 1}{x} = \dfrac{y - 2}{y} = \dfrac{z - 3}{z} \\ \dfrac{1}{x} + \dfrac{2}{y} + \dfrac{3}{z} = 2 \end{cases}\)
Đặt \(\dfrac{x - 1}{x}=\dfrac{y - 2}{y}=\dfrac{z - 3}{z}=k\)
=>\(\dfrac{1}{x}=\dfrac{2}{y}=\dfrac{3}{z}=1-k\)
\(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=2\)
=>1-k+1-k+1-k=2
=>3-3k=2
=>3k=1
=>k=1/3
=>\(\frac{1}{x}=\frac{2}{y}=\frac{3}{z}=1-\frac13=\frac23\)
=>x=3/2 và y=3 và z=9/2
=>a=3/2; b=1; c=5/6
\(c^2 + d^2 + 25 = 6c + 8d\)
=>\(c^2-6c+9+d^2-8d+16=0\)
=>\(\left(c-3\right)^2+\left(d-4\right)^2=0\)
=>c=3 và d=4
P=3*3+4*4-(a*3+b*4)=25-(3a+4b)
\((3a + 4b)^2 \le (3^2 + 4^2)(a^2 + b^2)\)
=>\((3a + 4b)^2 \le (9 + 16) \cdot 2 = 25 \cdot 2 = 50\)
=>\(-5\sqrt{2}\le3a+4b\le5\sqrt{2}\)
=>\(P \le 25 - (-5\sqrt{2}) = 25 + 5\sqrt{2}\)
Dấu '=' xảy ra khi \(\begin{cases} \dfrac{a}{3} = \dfrac{b}{4} < 0 \\ a^2 + b^2 = 2 \end{cases}\)
Đặt \(\frac{a}{3}=\frac{b}{4}=k\) (k<0)
=>a=3k; b=4k
\(a^2+b^2=2\)
=>\(\left(3k\right)^2+\left(4k\right)^2=2\)
=>\(25k^2=2\)
=>\(k^2=\frac{2}{25}\)
=>\(k=-\frac{\sqrt2}{5}\)
=>\(a=3k=-\frac{3\sqrt2}{5};b=4k=-\frac{4\sqrt2}{5}\)
Ta có đánh giá sau:
\(\dfrac{a^3}{\left(1-a\right)^2}\ge\dfrac{4a-1}{4}\)
Thật vậy, BĐT tương đương:
\(4a^3-\left(4a-1\right)\left(1-a\right)^2\ge0\)
\(\Leftrightarrow9a^2-6a+1\ge0\)
\(\Leftrightarrow\left(3a-1\right)^2\ge0\) (luôn đúng)
Tương tự: \(\dfrac{b^3}{\left(1-b\right)^2}\ge\dfrac{4b-1}{4}\) ; \(\dfrac{c^3}{\left(1-c\right)^2}\ge\dfrac{4c-1}{4}\)
Cộng vế:
\(P\ge\dfrac{4\left(a+b+c\right)-3}{4}=\dfrac{1}{4}\)
\(P_{min}=\dfrac{1}{4}\) khi \(a=b=c=\dfrac{1}{3}\)
a: Vì (P) đi qua A(0;1); B(1;2); C(3;-1) nên ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot0^2+b\cdot0+c=1\\a\cdot1^2+b\cdot1+c=2\\a\cdot3^2+b\cdot3+c=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=1\\a+b+1=2\\9a+3b+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=1\\a+b=1\\9a+3b=-2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=1\\9a+9b=9\\9a+3b=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=1\\6b=11\\a+b=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=1\\b=\dfrac{11}{6}\\a=1-\dfrac{11}{6}=-\dfrac{5}{6}\end{matrix}\right.\)
b: Vì (P) đi qua M(0;-1); N(1;0) và P(2;3) nên ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot0^2+b\cdot0+c=-1\\a\cdot1^2+b\cdot1+c=0\\a\cdot2^2+b\cdot2+c=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=-1\\a+b-1=0\\4a+2b-1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a+b=1\\4a+2b=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=-1\\a+b=1\\2a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\-a=-1\\a+b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-1\\a=1\\b=0\end{matrix}\right.\)
c: Vì (P) đi qua M(1;-2); N(0;4); P(2;1) nên ta có hệ phương trình:
\(\left\{{}\begin{matrix}a\cdot1^2+b\cdot1+c=-2\\a\cdot0^2+b\cdot0+c=4\\a\cdot2^2+b\cdot2+c=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a+b+c=-2\\c=4\\4a+2b+c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\a+b=-2-c=-6\\4a+2b=1-4=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=4\\4a+4b=-24\\4a+2b=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=4\\2b=-21\\a+b=-6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}c=4\\b=-\dfrac{21}{2}\\a=-6-b=-6+\dfrac{21}{2}=\dfrac{9}{2}\end{matrix}\right.\)
d: Hoành độ đỉnh là 2 nên -b/2a=2
=>b=-4a(1)
Thay x=3 và y=1 vào (P), ta được:
\(a\cdot3^2+b\cdot3+c=1\)
=>\(9a+3b+c=1\left(2\right)\)
Thay x=-1 và y=2 vào (P), ta được:
\(a\cdot\left(-1\right)^2+b\left(-1\right)+c=2\)
=>a-b+c=2(3)
Từ (1),(2),(3), ta có hệ phương trình:
\(\left\{{}\begin{matrix}b=-4a\\9a+3b+c=1\\a-b+c=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\9a-12a+c=1\\a+4a+c=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}b=-4a\\-3a+c=1\\5a+c=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-4a\\-8a=-1\\5a+c=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\dfrac{1}{8}\\b=-4\cdot\dfrac{1}{8}=-\dfrac{1}{2}\\c=2-5a=2-\dfrac{5}{8}=\dfrac{11}{8}\end{matrix}\right.\)
1) \(P=\left(a+2b+3c\right)\left(6a+3b+2c\right)\)
\(P=\left[a+2b+3\left(1-a-b\right)\right]+\left[6a+3b+2\left(1-a-b\right)\right]=\left(3-2a-b\right)\left(2+4a+b\right)=2\left(3a-2b-b\right)\left(1+2a+\dfrac{b}{2}\right)\)
Lợi dụng AM-GM, ta có:
\(P\le2\left(\dfrac{3-2a-b+1+2a+\dfrac{b}{2}}{2}\right)^2=2.\left(\dfrac{4-\dfrac{b}{2}}{2}\right)^2=8\)
MaxP=8 khi \(a=c=\dfrac{1}{2};b=0\)
Do \(a;b;c\in\left[0;1\right]\Rightarrow\left\{{}\begin{matrix}a-1\le0\\b-1\le0\\c-1\le0\end{matrix}\right.\) \(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\)
\(P=a+b+c-ab-bc-ca\)
\(=\left(a+b+c-ab-bc-ca+abc-1\right)+1-abc\)
\(=\left(a-1\right)\left(b-1\right)\left(c-1\right)+1-abc\)
\(\le1-abc\le1\)
\(P_{max}=1\) khi \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\) và các hoán vị