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\(VT=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{a^2}{b+c}\ge\frac{a^2}{\sqrt{2\left(b^2+c^2\right)}}+\frac{b^2}{\sqrt{2\left(c^2+a^2\right)}}+\frac{c^2}{\sqrt{2\left(c^2+a^2\right)}}\)
Đặt \(\left(\sqrt{b^2+c^2};\sqrt{c^2+a^2};\sqrt{a^2+b^2}\right)=\left(x;y;z\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=\frac{y^2+z^2-x^2}{2}\\b^2=\frac{x^2+z^2-y^2}{2}\\c^2=\frac{x^2+y^2-z^2}{2}\\x+y+z=\sqrt{2019}\end{matrix}\right.\) \(\Rightarrow VT\ge\frac{1}{\sqrt{8}}\left(\frac{y^2+z^2-x^2}{x}+\frac{x^2+z^2-y^2}{y}+\frac{x^2+y^2-z^2}{z}\right)\)
\(VT\ge\frac{1}{\sqrt{8}}\left(\frac{\left(y+z\right)^2}{2x}+\frac{\left(x+z\right)^2}{2y}+\frac{\left(x+y\right)^2}{2z}-\left(x+y+z\right)\right)\)
\(VT\ge\frac{1}{\sqrt{8}}\left[\frac{\left(2x+2y+2z\right)^2}{2\left(x+y+z\right)}-\left(x+y+z\right)\right]=\frac{x+y+z}{\sqrt{8}}=\sqrt{\frac{2019}{8}}\)
Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=\) nhiêu đó
Áp dụng bđt \(\frac{x^2}{m}+\frac{y^2}{n}+\frac{z^2}{p}\ge\frac{\left(x+y+z\right)^2}{m+n+p}\) ta có
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2}=a^2+b^2+c^2\)
Bài 1. Đặt \(a=\sqrt{x+3},b=\sqrt{x+7}\)
\(\Rightarrow a.b+6=3a+2b\) và \(b^2-a^2=4\)
Từ đó tính được a và b
Bài 2. \(\frac{2x-1}{x^2}+\frac{y-1}{y^2}+\frac{6z-9}{z^2}=\frac{9}{4}\)
\(\Leftrightarrow\frac{2}{x}-\frac{1}{x^2}+\frac{1}{y}-\frac{1}{y^2}+\frac{6}{z}-\frac{9}{z^2}-\frac{9}{4}=0\)
Đặt \(a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\)
Ta có \(2a-a^2+b-b^2+6c-9c^2-\frac{9}{4}=0\)
\(\Leftrightarrow-\left(a^2-2a+1\right)-\left(b^2-b+\frac{1}{4}\right)-\left(9c^2-6c+1\right)=0\)
\(\Leftrightarrow-\left(a-1\right)^2-\left(b-\frac{1}{2}\right)^2-\left(3c-1\right)^2=0\)
Áp dụng tính chất bất đẳng thức suy ra a = 1 , b = 1/2 , c = 1/3
Rồi từ đó tìm được x,y,z
1)
xét a+b+c = (a+b+c)(\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)) = \(\frac{a\left(a+b+c\right)}{b+c}+\frac{b\left(a+b+c\right)}{c+a}+\frac{c\left(a+b+c\right)}{a+b}=\)
\(\frac{a^2}{b+c}+\frac{a\left(b+c\right)}{b+c}+\frac{b^2}{a+c}+\frac{b\left(a+c\right)}{a+c}+\frac{c^2}{a+b}+\frac{c\left(a+b\right)}{a+b}=Q+a+b+c\)
<=> a+b+c =Q + a+b+c => Q=0
2) = (x+ y)2 + (x+ 1)2 +y(x+ 1) +x + y + 1 =0 <=> (x+ y)(x+ y+ 1) + (x+ 1)(x+ y+ 1) + 1= 0 <=> (x+ y+ 1)(2x+ y+ 1) = -1
=> \(\hept{\begin{cases}x+y+1=1\\2x+y+1=-1\end{cases}}\)hoặc \(\hept{\begin{cases}x+y+1=-1\\2x+y+1=1\end{cases}}\)<=> \(\hept{\begin{cases}x=-2\\y=2\end{cases}}\)hoặc \(\hept{\begin{cases}x=2\\y=-4\end{cases}}\)
Đặt \(\hept{\begin{cases}\left(b-c\right)\left(1+a\right)^2=m\\\left(c-a\right)\left(1+b\right)^2=n\\\left(a-b\right)\left(1+c\right)^2=p\end{cases}}\)
khi đó pt đã cho có dạng \(\frac{m}{x+a^2}+\frac{n}{x+b^2}+\frac{p}{x+c^2}=0\)
\(\Rightarrow m\left(x+a^2\right)\left(x+b^2\right)+n\left(x+a^2\right)\left(x+c^2\right)+p\left(x+b^2\right)\left(x+c^2\right)=0\)
\(\Rightarrow x^2\left(m+n+p\right)+x\left(m\left(a^2+b^2\right)+p\left(b^2+c^2\right)+n\left(c^2+a^2\right)\right)=0\)
Đến đây biện luận thôi ~~
Tớ làm hơi tắt đấy.
1. \(\left|2x-7\right|< x^2+2x+2\)
+) Xét \(x\ge\frac{7}{2}\):
Bpt \(\Leftrightarrow2x-7< x^2+2x+2\)
\(\Leftrightarrow x^2+9>0\) ( luôn đúng )
+) Xét \(x< \frac{7}{2}\):
Bpt \(\Leftrightarrow7-2x< x^2+2x+2\)
\(\Leftrightarrow x^2+4x-5>0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+5>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+5< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -5\end{matrix}\right.\)
Vậy...
2. \(a+b+c=0\Leftrightarrow a+c=-b\)
Ta có: \(P=\Sigma\frac{1}{b^2+\left(c-a\right)\left(c+a\right)}=\Sigma\frac{1}{b^2-b\left(c-a\right)}\)
\(=\Sigma\frac{1}{b\left(b-c+a\right)}=\Sigma\frac{1}{b\left(a+b+c-2c\right)}=\Sigma\frac{1}{-2bc}\)
\(=\frac{1}{2ab}+\frac{1}{2bc}+\frac{1}{2ca}=\frac{a+b+c}{2abc}=0\)
Vậy...