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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
2.
Từ giả thiết, ta có :
\(\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}+1-\frac{1}{1+d}\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự, ta cũng có :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{c.d.a}{\left(1+c\right)\left(1+d\right)\left(1+a\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chững minh rồi rút gọn ta được :
\(abcd\le\frac{1}{81}\left(đpcm\right)\)
2) Từ \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}\ge3.\)
\(\Rightarrow\frac{1}{1+a}\ge\left(1-\frac{1}{1+b}\right)+\left(1-\frac{1}{1+c}\right)+\left(1-\frac{1}{1+d}\right)\)
\(=\frac{b}{1+b}+\frac{c}{1+c}+\frac{d}{1+d}\ge3\sqrt[3]{\frac{bcd}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}.\)(BĐT AM-GM)
Tương tự :
\(\frac{1}{1+b}\ge3\sqrt[3]{\frac{acd}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\frac{1}{1+c}\ge3\sqrt[3]{\frac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\frac{1}{1+d}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}.\)
Từ đó suy ra:
\(\frac{1}{1+a}.\frac{1}{1+b}.\frac{1}{1+c}.\frac{1}{1+d}\ge3.3.3.3\sqrt[3]{\frac{\left(abcd\right)^3}{\left[\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)\right]^3}}\)
\(\Leftrightarrow\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}\ge\frac{81abcd}{\left(1+a\right)\left(1+b\right)\left(1+c\right)\left(1+d\right)}.\)
\(\Leftrightarrow81abcd\le1\Leftrightarrow abcd\le\frac{1}{81}\)
Dấu '=' xảy ra khi \(a=b=c=d=\frac{1}{3}.\)
3)Ta có: \(\left(\sqrt{a}+\sqrt{b}\right)^8=\left[\left(\sqrt{a}+\sqrt{b}\right)^2\right]^4=\left(a+b+2\sqrt{ab}\right)^4.\)(1)
Với \(a,b\ge0\),áp dụng BĐT AM-GM cho (a+b) và (\(2\sqrt{ab}\)) ta được
\(\left(a+b\right)+2\sqrt{ab}\ge2\sqrt{\left(a+b\right)2\sqrt{ab}}\)(2)
Từ (1) và (2) suy ra:
\(\left(\sqrt{a}+\sqrt{b}\right)^8\ge\left(2\sqrt{\left(a+b\right)2\sqrt{ab}}\right)^4\)
\(\Leftrightarrow\left(\sqrt{a}+\sqrt{b}\right)^8\ge64ab\left(a+b\right)^2.\)
Dấu '=' xảy ra khi \(a+b=2\sqrt{ab}\Leftrightarrow a=b\)
1) Với \(x\le\frac{2}{3}\Rightarrow2-3x\ge0\)
Khi đó ,áp dụng bất đẳng thức AM-GM cho 2 số ta được:
\(\left(2-3x\right)+\frac{9}{2-3x}\ge2\sqrt{\left(2-3x\right)\frac{9}{2-3x}}=2.3=6\)
\(\Leftrightarrow2+\left(2-3x\right)+\frac{9}{2-3x}\ge2+6\)
\(\Leftrightarrow4-3x+\frac{9}{2-3x}\ge8\)
Dấu '=' xảy ra khi \(2-3x=\frac{9}{2-3x}\Leftrightarrow\left(2-3x\right)^2=9\Leftrightarrow2-3x=3\Leftrightarrow x=-\frac{1}{3}\)( vì 2-3x>0)
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
Lời giải:
Ta có:
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\Rightarrow \frac{a}{b-c}=\frac{-b}{c-a}+\frac{-c}{a-b}\)
\(\Leftrightarrow \frac{a}{b-c}=\frac{-b(a-b)-c(c-a)}{(a-b)(c-a)}=\frac{b^2+ca-c^2-ab}{(a-b)(c-a)}\)
\(\Rightarrow \frac{a}{(b-c)^2}=\frac{b^2+ca-c^2-ab}{(a-b)(b-c)(c-a)}\)
Hoàn toàn tương tự:
\(\frac{b}{(c-a)^2}=\frac{c^2+ab-a^2-bc}{(a-b)(b-c)(c-a)}\)
\(\frac{c}{(a-b)^2}=\frac{a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}\)
Cộng theo vế các đẳng thức vừa thu được ta có:
\(\frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}=\frac{b^2+ac-c^2-ab+c^2+ab-a^2-bc+a^2+bc-b^2-ac}{(a-b)(b-c)(c-a)}=0\)
Ta có đpcm.
Ta có thể xét hiệu : \(\dfrac{a^2+b^2}{2}-\left(\dfrac{a+b}{2}\right)^2=\dfrac{2\left(a^2+b^2\right)}{4}-\dfrac{\left(a+b\right)^2}{4}\)
\(=\dfrac{2\left(a^2+b^2\right)-\left(a^2+2ab+b^2\right)}{4}\)
\(=\dfrac{1}{4}\left(a^2-2ab+b^2\right)=\dfrac{1}{4}\left(a-b\right)^2\)
Ta thấy : \(\left(a-b\right)^2\ge0\) nên \(\dfrac{1}{4}\left(a-b\right)^2\ge0\)
Hay là : \(\dfrac{a^2+b^2}{2}-\left(\dfrac{a+b}{2}\right)^2\ge0\)
Vậy \(\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
=> ĐPCM.
Từ \(\dfrac{a-\left(c-b\right)}{b-c}+\dfrac{b-\left(a-c\right)}{c-a}+\dfrac{c-\left(b-a\right)}{a-b}=3\)
\(=>\dfrac{a}{b-c}+1+\dfrac{b}{c-a}+1+\dfrac{c}{a-b}+1=3\)
\(=>\dfrac{a}{b-c}-\dfrac{b}{a-c}-\dfrac{c}{b-a}=0\)
\(=>\dfrac{a}{b-c}=\dfrac{b}{a-c}+\dfrac{c}{b-a}=\dfrac{b^2-ab+ac-c^2}{\left(c-a\right)\left(a-b\right)}\)
Nhân cả 2 vế với \(\dfrac{1}{b-c}\) ta được
\(\dfrac{a}{\left(b-c\right)^2}=\dfrac{b^2-ab+ac-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(1\right)\)
Tương tự ta có:
\(\dfrac{b}{\left(c-a\right)^2}=\dfrac{c^2-bc+bc-a^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(2\right)\)
\(\dfrac{c}{\left(a-b\right)^2}=\dfrac{a^2-ca+cb-c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left(3\right)\)
Cộng theo vế (1);(2);(3) ta có ĐPCM
CHÚC BẠN HỌC TỐT.........
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
\(\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2}\le a^2+b^2\)
\(\Leftrightarrow a^2+2ab+b^2\le2\left(a^2+b^2\right)\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( Luôn đúng )
\("="\Leftrightarrow a=b\)
bạn chứng minh sai rồi !