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Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Từ (a-1)(b-1)(c-1)>0 (*)
<=>(ab-b-a+1)(c-1)>0
<=> abc-ab-bc+b-ac+a+c-1>0
<=> a+b+c-ab-ac-bc>0
<=> a+b+c-\(\dfrac{abc}{c}-\dfrac{abc}{b}-\dfrac{abc}{a}\)>0
<=> a+b+c - \(\dfrac{1}{c}-\dfrac{1}{b}-\dfrac{1}{a}>0\)
<=> \(a+b+c>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( 1)
(1) đúng => (*) đúng
Bài 1: \(a+b\ge1\). cm \(a^4+b^4\ge\dfrac{1}{8}\)
ta có : \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)(BĐT bunyakovsky)
Áp dụng BĐt bunyakovsky 1 lần nữa:
\(a^4+b^4\ge\dfrac{1}{2}\left(a^2+b^2\right)^2\ge\dfrac{1}{2}.\dfrac{1}{4}=\dfrac{1}{8}\)
dấu = xảy ra khi \(a=b=\dfrac{1}{2}\)
Bài 2:
Áp dụng BĐT bunyakovsky dạng đa thức và phân thức:
\(\left(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\right)\left(a+b+c\right)\ge\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2\ge\left[\dfrac{\left(a+b+c\right)^2}{a+b+c}\right]^2=\left(a+b+c\right)^2\)
do đó \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\)
dấu = xảy ra khi a=b=c
Bài 1:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2=1\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge1\Rightarrow a^2+b^2\ge\dfrac{1}{2}\)
Lại theo Cauchy-Schwarz lần nữa:
\(\left[\left(1^2\right)^2+\left(1^2\right)^2\right]\left[\left(a^2\right)^2+\left(b^2\right)^2\right]\ge\left(a^2+b^2\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow2\left(a^4+b^4\right)\ge\dfrac{1}{4}\Leftrightarrow a^4+b^4\ge\dfrac{1}{8}\)
Đẳng thức xảy ra khi \(a=b=\dfrac{1}{2}\)
Bài 2:
Trước tiên ta chứng minh \(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\)
Ta chứng minh bổ đề: \(\dfrac{a^3}{b^2}\ge\dfrac{a^2}{b}+a-b\)
\(\Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng)
Viết các BĐT tương tự và cộng lại
\(\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge\dfrac{a^2}{b}+a-b+\dfrac{b^2}{c}+b-c+\dfrac{c^2}{a}+c-a=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\left(1\right)\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\left(2\right)\)
Từ \((1);(2)\) ta thu được ĐPCM
Ta có :
\(\dfrac{a^2}{a^2+3}>\dfrac{a^2}{a^2+b^2+c^2+4}\)
\(\dfrac{b^2}{b^2+2}>\dfrac{b^2}{a^2+b^2+c^2+4}\)
\(\dfrac{c^2}{c^2+1}>\dfrac{c^2}{a^2+b^2+c^2+4}\)
\(\dfrac{4}{a^2+4+c^2}\ge\dfrac{4}{a^2+b^2+c^2+4}\)
Cộng vế với vế lại ta được :
\(P>\dfrac{a^2+b^2+c^2+4}{a^2+b^2+c^2+4}=1\) (đpcm)
Áp dụng bđt AM-GM: \(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2a}}=2\sqrt{\dfrac{1}{b^2}}=\dfrac{2}{b}\) \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge2\sqrt{\dfrac{b}{c^2b}}=2\sqrt{\dfrac{1}{c^2}}=\dfrac{2}{c}\) \(\dfrac{c}{a^2}+\dfrac{1}{c}\ge2\sqrt{\dfrac{c}{a^2c}}=2\sqrt{\dfrac{1}{a^2}}=\dfrac{2}{a}\) Cộng theo vế: \(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\Leftrightarrow\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)Dấu "=" xảy ra khi: \(a=b=c\)
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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
C1:Áp dụng Bất đẳng thức AM-GM ta có:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1^2}{a+b}+\dfrac{1^2}{b+c}+\dfrac{1^2}{c+a}\ge\)
\(\ge\dfrac{\left(1+1+1\right)^2}{a+b+b+c+c+a}=\dfrac{9}{2\left(a+b+c\right)}\)
\(\Rightarrow A=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}\)Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
C2: Khai triển
\(A=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\)
\(=1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}\) (bn tự khai triển đầy đủ nha)
Áp dụng BĐT Nesbitt ta có:
\(A=\left(1+1+1\right)+\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\ge\)
\(\left(1+1+1\right)+\dfrac{3}{2}=\dfrac{9}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)