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choa,b,c >0.CMR:\(\dfrac{11a^3-b^3}{4a^2+ab}+\dfrac{11b^3-c^3}{4b^2+bc}+\dfrac{11c^3-a^3}{4c^2+ac}\)
Đã thấy. Sửa đề: \(\sum\dfrac{11a^3-b^3}{4a^2+ab}\le2\left(a+b+c\right)\)
\(\sum\dfrac{11a^3-b^3}{4a^2+ab}=\sum\dfrac{12a^3-\left(a^3+b^3\right)}{4a^2+ab}=\sum\dfrac{12a^3-\left(a+b\right)\left(\left(a-b\right)^2+ab\right)}{4a^2+ab}\)
\(\le\sum\dfrac{12a^3-ab\left(a+b\right)}{4a^2+ab}=\sum\dfrac{a\left(3a-b\right)\left(4a+b\right)}{a\left(4a+b\right)}\)
\(=\sum\left(3a-b\right)=2\left(a+b+c\right)\)
Đề bài: Cho \(a,b,c>0\). CMR \( \frac{11b^3-a^3}{ab+4b^2} + \frac{11c^3-b^3}{bc+4c^2} + \frac{11a^3-c^3}{ac+4a^2} \leq 2(a+b+c)\)
Bài giải
Ta chứng minh bổ đề \(\dfrac{11b^3-a^3}{4b^2+ab}\le3b-a\)
Thật vậy \(11b^3-a^3\le\left(ab+4b^2\right)\left(3b-a\right)\Leftrightarrow11b^3-a^3\le-a^2b-ab^2+12b^3\)
\(\Leftrightarrow a^3-a^2b-ab^2+b^3\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (đúng)
Tương tự cho2 BĐT còn lại ta cũng có:
\(\dfrac{11c^3-b^3}{4c^2+bc}\le3c-b;\dfrac{11a^3-c^3}{4a^2+ac}\le3a-c\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\left(3b-a\right)+\left(3c-b\right)+\left(3a-c\right)=2\left(a+b+c\right)=VP\)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
Lời giải:
Đặt \(\frac{ab}{c}=x; \frac{bc}{a}=y; \frac{ca}{b}=z\Rightarrow a^2=xz; b^2=xy; c^2=yz\)
Bài toán trở thành: Cho $x,y,z>0$ thỏa mãn \(xy+yz+xz=3\)
Chứng minh \(x+y+z\geq 3\)
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Theo hệ quả quen thuộc của BĐT AM-GM:
\(x^2+y^2+z^2\geq xy+yz+xz\)
\(\Rightarrow x^2+y^2+z^2+2(xy+yz+xz)\geq 3(xy+yz+xz)\)
\(\Leftrightarrow (x+y+z)^2\geq 3(xy+yz+xz)=9\)
\(\Rightarrow x+y+z\geq 3\)
Ta có đpcm
Dấu "=" xảy ra khi $x=y=z=1$ hay $a=b=c=1$
\(A\le\frac{1}{27}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^3\)
Mặt khác :
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3\left[4\left(a+b+c\right)+3\right]}\)
\(=3\sqrt{5}\)
\(\Rightarrow A\le\frac{1}{27}\left(3\sqrt{5}\right)^3=5\sqrt{5}\)
Dấu " = " xảy ra khi \(a=b=c=1\)
Áp dụng BĐT cauchy-schwarz:
\(\sum\dfrac{a^4b}{2a+b}=\sum\dfrac{a^4b^2}{2ab+b^2}\ge\dfrac{\left(a^2b+b^2c+c^2a\right)^2}{\left(a+b+c\right)^2}\)
giờ ta chỉ cần có:\(a^2b+b^2c+c^2a\ge a+b+c\)
Áp dụng AM-GM:
\(a^2b+\dfrac{1}{b}\ge2a\)..tương tự ,ta suy ra:
\(a^2b+b^2c+c^2a\ge2\left(a+b+c\right)-\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)(*)
Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le3\)
Dễ dàng suy ra được \(a+b+c\ge3\) ( từ BĐT \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\))
theo đó thì \(a+b+c\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Nên từ (*) ta có đpcm.
Dấu = xảy ra khi a=b=c=1
\(A=3\left(ab+bc+ca\right)+\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{4}\left(b-c\right)^2+\dfrac{1}{8}\left(c-a\right)^2\\ =3\left(ab+bc+ca\right)+\dfrac{\left(a-b\right)^2}{2}+\dfrac{\left(b-c\right)^2}{4}+\dfrac{\left(c-a\right)^2}{8}\)
Áp dụng BDT: Cô-si dạng Engel:
\(\Rightarrow A=3\left(ab+bc+ca\right)+\dfrac{\left(a-b\right)^2}{2}+\dfrac{\left(b-c\right)^2}{4}+\dfrac{\left(c-a\right)^2}{8}\ge3\left(ab+bc+ca\right)+\dfrac{\left(a-b+b-c+c-a\right)^2}{2+4+8}=3\left(ab+bc+ca\right)\left(1\right)\)
\(\text{Ta lại có: }ab+bc+ac\le a^2+b^2+c^2\\ \Leftrightarrow ab+bc+ac+2\left(ab+bc+ac\right)\le a^2+b^2+c^2+2\left(ab+bc+ac\right)\\ \Leftrightarrow3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2=3^2=9\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\Rightarrow A\le9\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}a=b=c\\a+b+c=3\\\dfrac{a-b}{2}+\dfrac{b-c}{4}+\dfrac{c-a}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\Leftrightarrow a=b=c=1\)
Vậy \(A_{Max}=9\) khi \(a=b=c=1\)
a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)
\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)
\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)
b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)
\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)
\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)
\(ab+bc+ca=3\Rightarrow\left\{{}\begin{matrix}a+b+c\ge3\\abc\le1\end{matrix}\right.\)
Ta sẽ chứng minh \(P\le\dfrac{3}{8}\)
\(P\le\dfrac{a}{6a+2}+\dfrac{b}{6b+2}+\dfrac{c}{6c+2}\) nên chỉ cần chứng minh: \(\dfrac{a}{3a+1}+\dfrac{b}{3b+1}+\dfrac{c}{3c+1}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{3a+1}+\dfrac{1}{3b+1}+\dfrac{1}{3c+1}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{\left(3a+1\right)\left(3b+1\right)+\left(3b+1\right)\left(3c+1\right)+\left(3c+1\right)\left(3a+1\right)}{\left(3a+1\right)\left(3b+1\right)\left(3c+1\right)}\ge\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{6\left(a+b+c\right)+30}{27abc+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)
\(\Rightarrow\dfrac{6\left(a+b+c\right)+30}{27+3\left(a+b+c\right)+28}\ge\dfrac{3}{4}\)
\(\Leftrightarrow24\left(a+b+c\right)+120\ge165+9\left(a+b+c\right)\)
\(\Leftrightarrow a+b+c\ge3\) (đúng)