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Ta có:
\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{1^2}{a^3\left(b+c\right)}+\dfrac{1^2}{b^3\left(c+a\right)}+\dfrac{1^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{a^2b^2c^2}{a^3\left(b+c\right)}+\dfrac{a^2b^2c^2}{b^3\left(c+a\right)}+\dfrac{a^2b^2c^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
\(\dfrac{b^2c^2}{a\left(c+b\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{3}{2}\)
Áp dụng BĐT Svacxo ta có:
\(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)^2}{a\left(b+c\right)+b\left(a+c\right)+c\left(a+b\right)}\) \(\dfrac{b^2c^2}{a\left(b+c\right)}+\dfrac{a^2c^2}{b\left(c+a\right)}+\dfrac{a^2b^2}{c\left(a+b\right)}\ge\dfrac{\left(ab+bc+ca\right)}{2}\) (1)
Chứng minh: \(\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\Leftrightarrow ab+bc+ca\ge3\)
Áp dụng BĐT Cosi ta có:
\(ab+bc+ca\ge3\sqrt[3]{ab.bc.ca}\)
\(ab+bc+ca\ge3\) (2)
Từ (1) và (2)
=> ĐPCM
a)\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
=\(\dfrac{a}{a}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{c}+\dfrac{c}{a}+\dfrac{c}{b}\)
=\(1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)
=3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)
áp dụng BĐT cô si ta có
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)
⇔ \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
cmtt ta có \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\); \(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
=> 3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge9\)
=> \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)
a)Áp dụng bđt AM-GM cho 3 số không âm ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
TT\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
Nhân vế theo vế ta có:\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\left(đpcm\right)\)
b)\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\)
Svac-xo:
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Lại có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(tự cm)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Lời giải:
Ta có:
\(\text{VT}=\frac{a+c+2c}{a+b}+\frac{a+b+2b}{a+c}+\frac{2a}{b+c}\)
\(=\left(\frac{a+c}{a+b}+\frac{a+b}{a+c}\right)+2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)\)
Áp dụng BĐT AM-GM: \(\frac{a+c}{a+b}+\frac{a+b}{a+c}\geq 2\)
Và:
\(2\left(\frac{c}{a+b}+\frac{b}{a+c}+\frac{a}{b+c}\right)=2\left(\frac{c+a+b}{a+b}+\frac{b+a+c}{a+c}+\frac{a+b+c}{b+c}-3\right)\)
\(=2(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-6\)
\(=[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-6\)
\(\geq 3\sqrt[3]{(a+b)(b+c)(c+a)}.3\sqrt[3]{\frac{1}{(a+b)(b+c)(c+a)}}-6=9-6=3\)
Do đó:
\(\text{VT}\geq 2+3=5\)
Ta có đpcm
Dấu bằng xảy ra khi $a=b=c$
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
ab−c−ba−c−cb−a=0=>ab−c−ba−c−cb−a=0
=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)=>ab−c=ba−c+cb−a=b2−ab+ac−c2(c−a)(a−b)
Nhân cả 2 vế với 1b−c1b−c ta được
a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)a(b−c)2=b2−ab+ac−c2(a−b)(b−c)(c−a)(1)
Tương tự ta có:
b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)b(c−a)2=c2−bc+bc−a2(a−b)(b−c)(c−a)(2)
c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)c(a−b)2=a2−ca+cb−c2(a−b)(b−c)(c−a)(3)
Cộng theo vế (1);(2);(3) ta có ĐPCM
BĐT cần chứng minh tương đương với:
\(\left(\dfrac{a}{b^2}-\dfrac{2}{b}+\dfrac{1}{a}\right)+\left(\dfrac{b}{a^2}-\dfrac{2}{a}+\dfrac{1}{b}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-\dfrac{16}{a+b}\)
\(\Leftrightarrow\left(a+b\right)\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2\ge\dfrac{4\left(a-b\right)^2}{ab\left(a+b\right)}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a-b\right)^2}{a^2b^2}\ge\dfrac{4\left(a-b\right)^2}{ab\left(a+b\right)}\).
\(\Leftrightarrow\left(a-b\right)^2\left[\dfrac{a+b}{a^2b^2}-\dfrac{4}{ab\left(a+b\right)}\right]\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^4}{a^2b^2\left(a+b\right)}\ge0\) (luôn đúng).
`a/b^2+b/a^2+16/(a+b)>=5(1/a+1/b)`
`<=>a/b^2-1/b+b^2-1/a+4(4/(a+b)-1/a-1/b)=0`
`<=>(a-b)/b^2+(b-a)/a^2+4((4ab-(a+b)^2)/(ab(a+b)))>=0`
`<=>(a^2(a-b)-b^2(a-b))/(a^2b^2)-(4(a-b)^2)/(ab(a+b))>=0`
`<=>(a-b)^2[(a+b)^2-4ab]>=0`
`<=>(a-b)^2(a^2-2ab+b^2)>=0`
`<=>(a-b)^2(a-b)^2>=0`
`<=>(a-b)^4>=0` luôn đúng.
Dấu "=" xảy ra khi `a=b`