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Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
\(\dfrac{b+c-a}{2a}+\dfrac{a-b+c}{2b}+\dfrac{a+b-c}{2c}\ge\dfrac{3}{2}\)
Ta có: \(\dfrac{b+c-a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}\)
Viết lại BĐT cần chứng minh như sau:
\(\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}+\dfrac{a}{2b}-\dfrac{1}{2}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{1}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{3}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{b}{2a}+\dfrac{a}{2b}=\dfrac{1}{2}\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\cdot\dfrac{1}{2}=1\)
\(\dfrac{c}{2a}+\dfrac{a}{2c}=\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{c}{a}+\dfrac{a}{c}}=\dfrac{1}{2}\cdot2=1\)
\(\dfrac{b}{2c}+\dfrac{c}{2b}=\dfrac{1}{2}\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{c}\cdot\dfrac{c}{b}}=\dfrac{1}{2}\cdot2=1\)
Cộng theo vế 3 BĐT trên ta có:
\(\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}\ge3\)
\(\Rightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge3-3=0\)
BĐT đúng nên ta có ĐPCM
á mk xl nhá mk ko đọc kĩ đề mk làm nhầm rùi bài mk làm là tìm GTNN nhá bạn ( mất công quá
)
ta có A= a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
= \(\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c}{4}+\dfrac{3c}{4}+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)
=\(\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)
vì a,b,c >0 ===> \(\dfrac{3a}{4}>0,\dfrac{3}{a}>0,\dfrac{b}{2}>0,\dfrac{9}{2b}>0,\dfrac{c}{4}>0,\dfrac{4}{c}>0\)
áp dụng BĐT côsi cho các cặp số dương ta đc:
\(\dfrac{3a}{4}+\dfrac{3}{a}>=2.\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}=3\)
\(\dfrac{b}{2}+\dfrac{9}{2b}>=3\)(làm như trên nhá)
\(\dfrac{c}{4}+\dfrac{4}{c}>=2\)
===> \(\dfrac{3a}{4}+\dfrac{3}{a}+\dfrac{b}{2}+\dfrac{9}{2b}+\dfrac{c}{4}+\dfrac{4}{c}>=8\left(1\right)\)
có: \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}=\dfrac{a+2b+3c}{4}\)
mà a+2b+3c >= 20
===> \(\dfrac{a+2b+3c}{4}>=\dfrac{20}{4}=5\)
===> \(\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}>=5\left(2\right)\)
từ (1) và(2)===> a+b+c+\(\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}>=13\)
===> A >= 13
Dấu ''='' xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Vậy Min A=13 <=>\(\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(B=\frac{1}{(a+2b)(a+2c)}+\frac{1}{(b+2a)(b+2c)}+\frac{1}{(c+2a)(c+2b)}\)
\(\geq \frac{9}{(a+2b)(a+2c)+(b+2a)(b+2c)+(c+2a)(c+2b)}\)
\(\Leftrightarrow B\geq \frac{9}{(a^2+2ac+2ab+4bc)+(b^2+2bc+2ab+4ac)+(c^2+2bc+2ac+4ab)}\)
\(\Leftrightarrow B\geq \frac{9}{a^2+b^2+c^2+8(ab+bc+ac)}=\frac{9}{(a+b+c)^2+6(ab+bc+ac)}(*)\)
Theo hệ quả quen thuộc của BĐT Cô-si:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Rightarrow 2(a+b+c)^2\geq 6(ab+bc+ac)(**)\)
Từ \((*); (**)\Rightarrow B\geq \frac{9}{(a+b+c)^2+2(a+b+c)^2}=\frac{3}{(a+b+c)^2}\geq \frac{3}{3^2}=\frac{1}{3}\)
(do \(a+b+c\leq 3)\)
Do đó: \(B_{\min}=\frac{1}{3}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{b}=\dfrac{2}{c}-\dfrac{1}{a}=\dfrac{2a-c}{ac}\\\dfrac{1}{a}=\dfrac{2}{c}-\dfrac{1}{b}=\dfrac{2b-c}{bc}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a-c=\dfrac{ac}{b}\\2b-c=\dfrac{bc}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+c}{2a-c}=\dfrac{b\left(a+c\right)}{ac}=\dfrac{ab}{ac}+\dfrac{bc}{ac}=\dfrac{b}{c}+\dfrac{b}{a}\\\dfrac{b+c}{2b-c}=\dfrac{a\left(b+c\right)}{bc}=\dfrac{ab}{bc}+\dfrac{ac}{bc}=\dfrac{a}{c}+\dfrac{a}{b}\end{matrix}\right.\)
Áp dụng bđt Cosi cho 2 số sương ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{2}{c}=0\Leftrightarrow\dfrac{2}{c}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Leftrightarrow\dfrac{a+b}{c}\ge2\)(áp dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\))
Ta có: \(\dfrac{a+c}{2a-c}+\dfrac{b+c}{2b-c}=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\dfrac{a+b}{c}\ge2+2=4\)
Dấu "=" xawy ra khi và chỉ khi a=b=c
Lời giải:
Biến đổi $A$ :
\(A=a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}=\frac{1}{4}(a+2b+3c)+\left(\frac{3a}{4}+\frac{3}{a}\right)+\left (\frac{b}{2}+\frac{9}{2b}\right)+\left (\frac{c}{4}+\frac{4}{c}\right)\)
Ta có: \(\frac{1}{4}(a+2b+3c)\geq \frac{20}{4}=5\)
Áp dụng BĐT AM-GM: \(\left\{\begin{matrix} \frac{3a}{4}+\frac{3}{a}\geq 3\\ \frac{b}{2}+\frac{9}{2b}\geq 3\\ \frac{c}{4}+\frac{4}{c}\geq 2\end{matrix}\right.\)
Do đó \(A\geq 5+3+3+2=13\) hay \(A_{\min}=13\)
Dấu bằng xảy ra khi \(\left\{\begin{matrix} a=2\\ b=3\\ c=4\end{matrix}\right.\)
Mấu chốt của bài toán là cách tìm điểm rơi.