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\(6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)\)
\(=6\left(x^2+y^2+z^2\right)+12\left(xy+yz+xz\right)+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\right)-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+2\left(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{2z+x+y}\right)-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+2.\dfrac{\left(1+1+1\right)^2}{2x+y+z+x+2y+z+2z+x+y}-2\left(xy+yz+xz\right)\)
\(=6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-2\left(xy+yz+xz\right)\)
\(\ge6\left(x+y+z\right)^2+\dfrac{18}{4\left(x+y+z\right)}-\dfrac{2}{3}\left(x+y+z\right)^2\)
\(=6.\left(\dfrac{3}{4}\right)^2+\dfrac{18}{4.\dfrac{3}{4}}-\dfrac{2}{3}.\left(\dfrac{3}{4}\right)^2=9\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
a) ab+bc+ca\(\le\dfrac{\left(a+c+b\right)^2}{3}\)
\(\Leftrightarrow3ab+3bc+3ac\le a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow ab+bc+ac\le a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ca\le2a^2+2b^2+2c^2\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng \(\forall a,b,c\)
Chính bài của em:
Cho \(a,b,c\ge1\). CMR: \(a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}... - Hoc24
do \(a,b,c\ge1\)\(=>\left\{{}\begin{matrix}b+c\ge2\\c+a\ge2\\a+b\ge2\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}a\left(b+c\right)\ge2a\\b\left(c+a\right)\ge2b\\c\left(a+b\right)\ge2c\end{matrix}\right.\)
\(=>\) biểu thức đề bài cho\(\ge2\left(a+b+c+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
\(2\left(1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)=9\)
dấu= xảy ra<=>a=b=c=1
Lời giải:
BĐT cần CM tương đương với:
\(\left[\frac{(a+b)(1-ab)}{(a^2+1)(b^2+1)}\right]^2\leq \frac{1}{4}\)
Đặt $a+b=x; ab=y$ thì BĐT \(\Leftrightarrow \left(\frac{x(1-y)}{y^2+x^2-2y+1}\right)^2=\left(\frac{x(y-1)}{x^2+(y-1)^2}\right)^2\leq \frac{1}{4}\)
Điều này luôn đúng vì theo BĐT AM-GM:
\([x^2+(y-1)^2]^2=x^4+(y-1)^4+2x^2(y-1)^2\geq 2x^2(y-1)^2+2x^2(y-1)^2=[2x(y-1)]^2\)
\(\Rightarrow \frac{[x(y-1)]^2}{[x^2+(y-1)^2]^2}\leq \frac{[x(y-1)]^2}{[2x(y-1)]^2}=\frac{1}{4}\)
Fix đề: Cho a,b,c không âm. Chứng minh \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(b-c\right)^2}+\dfrac{1}{\left(c-a\right)^2}\ge\dfrac{4}{ab+bc+ca}\)
Dự đoán điểm rơi sẽ có 1 số bằng 0.
Giả sử \(c=min\left\{a,b,c\right\}\) ( c là số nhỏ nhất trong 3 số) thì \(c\ge0\)
do đó \(ab+bc+ca\ge ab\) và \(\dfrac{1}{\left(b-c\right)^2}\ge\dfrac{1}{b^2};\dfrac{1}{\left(c-a\right)^2}=\dfrac{1}{\left(a-c\right)^2}\ge\dfrac{1}{a^2}\)
BDT cần chứng minh tương đương
\(ab\left[\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\right]\ge4\)
\(\Leftrightarrow\dfrac{ab}{\left(a-b\right)^2}+\dfrac{a^2+b^2}{ab}\ge4\)
\(\Leftrightarrow\dfrac{ab}{\left(a-b\right)^2}+\dfrac{\left(a-b\right)^2}{ab}+2\ge4\)
BĐT trên hiển nhiên đúng theo AM-GM.
Do đó ta có đpcm. Dấu = xảy ra khi c=0 , \(\left(a-b\right)^2=a^2b^2\) ( và các hoán vị )
Lời giải:
Từ \(ab+a+b=1\) suy ra :
\(a^2+1=a^2+ab+a+b=a(a+b)+(a+b)=(a+1)(a+b)\)
\(b^2+1=b^2+ab+a+b=b(b+a)+(a+b)=(b+1)(a+b)\)
Do đó:
\(\frac{a}{a^2+1}+\frac{b}{b^2+1}=\frac{a}{(a+1)(a+b)}+\frac{b}{(b+1)(a+b)}=\frac{a(b+1)+b(a+1)}{(a+1)(b+1)(a+b)}\)
\(=\frac{ab+(ab+a+b)}{(a+1)(b+1)(a+b)}=\frac{ab+1}{(a+1)(b+1)(a+b)}(*)\)
Và:
\(\frac{ab+1}{\sqrt{2(a^2+1)(b^2+1)}}=\frac{ab+1}{\sqrt{2(a+1)(a+b)(b+1)(a+b)}}=\frac{ab+1}{\sqrt{(ab+a+b+1)(a+1)(b+1)(a+b)^2}}\)
\(=\frac{ab+1}{\sqrt{(a+1)(b+1)(a+1)(b+1)(a+b)^2}}=\frac{ab+1}{(a+1)(b+1)(a+b)}(**)\)
Từ $(*); (**)$ ta có đpcm.
\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)
Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)
\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(P\ge\dfrac{3}{2}+\dfrac{15}{32}\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=9\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)