Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
12. Ta có \(ab\le\frac{a^2+b^2}{2}\)
=> \(a^2-ab+3b^2+1\ge\frac{a^2}{2}+\frac{5}{2}b^2+1\)
Lại có \(\left(\frac{a^2}{2}+\frac{5}{2}b^2+1\right)\left(\frac{1}{2}+\frac{5}{2}+1\right)\ge\left(\frac{a}{2}+\frac{5}{2}b+1\right)^2\)
=> \(\sqrt{a^2-ab+3b^2+1}\ge\frac{a}{4}+\frac{5b}{4}+\frac{1}{2}\)
=> \(\frac{1}{\sqrt{a^2-ab+3b^2+1}}\le\frac{4}{a+b+b+b+b+b+1+1}\le\frac{4}{64}.\left(\frac{1}{a}+\frac{5}{b}+2\right)\)
Khi đó
\(P\le\frac{1}{16}\left(6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+6\right)\le\frac{3}{2}\)
Dấu bằng xảy ra khi a=b=c=1
Vậy \(MaxP=\frac{3}{2}\)khi a=b=c=1
13. Ta có \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le1\)
\(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)( BĐT cosi)
=> \(1\ge\frac{9}{a+b+c+3}\)
=> \(a+b+c\ge6\)
Ta có \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
=> \(\frac{a^3-b^3}{a^2+ab+b^2}=a-b\)
Tương tự \(\frac{b^3-c^3}{b^2+bc+c^2}=b-c\),,\(\frac{c^3-a^2}{c^2+ac+a^2}=c-a\)
Cộng 3 BT trên ta có
\(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ac+c^2}=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{c^2+bc+b^2}+\frac{a^3}{a^2+ac+c^2}\)
Khi đó \(2P=\frac{a^3+b^3}{a^2+ab+b^2}+...\)
=> \(2P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}+....\)
Xét \(\frac{a^2-ab+b^2}{a^2+ab+b^2}\ge\frac{1}{3}\)
<=> \(3\left(a^2-ab+b^2\right)\ge a^2+ab+b^2\)
<=> \(a^2+b^2\ge2ab\)(luôn đúng )
=> \(2P\ge\frac{1}{3}\left(a+b+b+c+a+c\right)=\frac{2}{3}.\left(a+b+c\right)\ge4\)
=> \(P\ge2\)
Vậy \(MinP=2\)khi a=b=c=2
Lưu ý : Chỗ .... là tương tự
\(A=\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}+3\\ =\left(\dfrac{1}{2a-a^2}+\dfrac{1}{2b-b^2}+\dfrac{1}{2c-c^2}\right)+3\\ \overset{AM-GM}{\ge}\dfrac{9}{2a-a^2+2b-b^2+2c-c^2}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ =\dfrac{9}{\left(2a+2b+2c\right)-\left(a^2+b^2+c^2\right)}+3\\ \ge\dfrac{9}{2\left(a+b+c\right)-\dfrac{\left(a+b+c\right)^2}{3}}+3\\ =\dfrac{9}{2\cdot1-\dfrac{1}{3}}+3=\dfrac{42}{5}\)
Dấu \("="\) xảy ra khi : \(\left\{{}\begin{matrix}2a-a^2=2b-b^2=2c-c^2\\a=b=c\\a+b+c=1\end{matrix}\right.\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Vậy \(A_{Min}=\dfrac{42}{5}\) khi \(a=b=c=\dfrac{1}{3}\)
+ \(2a+b+c=\left(a+b\right)+\left(a+c\right)\)
\(\ge2\sqrt{\left(a+b\right)\left(a+c\right)}\) ( theo AM-GM )
\(\Rightarrow\left(2a+b+c\right)^2\ge4\left(a+b\right)\left(a+c\right)\)
\(\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{4\left(a+b\right)\left(a+c\right)}\)
Dấu "=" xảy ra \(\Leftrightarrow b=c\)
+ Tương tự : \(\frac{1}{\left(2b+c+a\right)^2}\le\frac{1}{4\left(a+b\right)\left(b+c\right)}\). Dấu "=" xảy ra <=> a = c
\(\frac{1}{\left(2c+a+b\right)^2}\le\frac{1}{4\left(a+c\right)\left(b+c\right)}\). Dấu "=" xảy ra \(\Leftrightarrow a=b\)
Do đó : \(P\le\frac{1}{4}\left(\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(b+c\right)}+\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\)
\(\Rightarrow P\le\frac{1}{2}\cdot\frac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}\)\(=8abc\)
\(\Rightarrow P\le\frac{a+b+c}{16abc}\)
+ \(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\). Dấu :=" xảy ra \(\Leftrightarrow a=b\)
\(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\). Dấu "=" xảy ra <=> b = c
\(\frac{1}{c^2}+\frac{1}{a^2}\ge\frac{2}{ca}\). Dấu "=" xảy ra <=> c = a
\(\Rightarrow2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(\Rightarrow3\ge\frac{a+b+c}{abc}\) \(\Rightarrow a+b+c\le3abc\)
\(\Rightarrow P\le\frac{3abc}{16abc}=\frac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Lời giải:
Đặt \(\left(\frac{ab}{c}, \frac{bc}{a}, \frac{ca}{b}\right)=(x,y,z)\)
Khi đó: \(xy=b^2; yz=c^2; xz=a^2\). Bài toán trở về dạng:
Cho $x,y,z>0$ thỏa mãn: \(xy+yz+xz=1\)
Tìm GTNN của \(P=x+y+z\)
Thật vậy: Ta đã biết một BĐT quen thuộc theo AM-GM là:
\((x+y+z)^2\geq 3(xy+yz+xz)\)
\(\Rightarrow x+y+z\geq \sqrt{3(xy+yz+xz)}=\sqrt{3}\)
Vậy \(P_{\min}=\sqrt{3}\)
Dấu bằng xảy ra khi \(x=y=z\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vì vai trò của a,b,c là như nhau, giả sử
\(a\ge c\ge b>0\)
Ta có
\(a+b-c< a\)
\(\Leftrightarrow b-c\le0\) ( đúng với gt )
\(\Rightarrow a+b-c< a\)
\(\Leftrightarrow\left(a+b-c\right)^2< a^2\)
\(\Leftrightarrow\dfrac{1}{\left(a+b-c\right)^2}\ge\dfrac{1}{a^2}\)
CMTT :
\(\dfrac{1}{\left(b+c-a\right)^2}\ge\dfrac{1}{b^2};\dfrac{1}{\left(c+a-b\right)^2}\ge\dfrac{1}{c^2}\)
Cộng vế với vế 3 BĐT trên , được
\(\dfrac{1}{\left(a+b-c\right)^2}+\dfrac{1}{\left(b+c-a\right)^2}+\dfrac{1}{\left(c+a-b\right)^2}\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
