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đăng từng câu 1 thôi, nhiều nhất là 3 câu/ 1 lần hỏi vì đâu có giới hạn số lần hỏi
a.
Xét hiệu:
\(a^3+b^3-ab\left(a+b\right)=\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\)
\(=a^2-ab+b^2-ab=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\ge0\)
=> BĐT luôn đúng
b.
Xét hiệu:
\(a^4+b^4-a^3b-ab^3=\left(a^4-a^3b\right)-\left(b^4-ab^3\right)\)
\(=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a^3-b^3\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a-b\right)\)
\(=\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
=> BĐT luôn đúng
a)
\(a^3+b^3\ge ab\left(a+b\right)\forall a,b>0\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Rightarrow a^2-ab+b^2\ge ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)
\(\Rightarrowđpcm\)
b)
\(a^4+b^4\ge a^3b+ab^3\)
\(\Leftrightarrow a^4-ab^3+b^4-a^3b\ge0\)
\(\Leftrightarrow a\left(a^3-b^3\right)-b\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Rightarrowđpcm\)
c)
\(\left(a+1\right)\left(b+1\right)\ge\left(\sqrt{ab}+1\right)^2\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)-\left(\sqrt{ab}+1\right)^2\ge0\)
\(\Leftrightarrow1+b+a+ab-ab-2\sqrt{ab}-1\ge0\)
\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)
Dấu bằng xảy ra khi \(a=b\)
d)
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\)
Áp dụng bất đẳng thức AM-GM ta được
\(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}\)
\(\Leftrightarrow\dfrac{a^3}{b}+ab\ge2a^2\)
Tương tự ta được
\(\dfrac{b^3}{c}+bc\ge2b^2,\dfrac{c^3}{a}+ac\ge2c^2\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ac\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ac\right)\)
Mặt khác ta có:\(a^2+b^2+c^2\ge ab+bc+ac\) (hệ quả bất đẳng thức AM-GM)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ac\left(đpcm\right)\)
Dấu bằng xảy ra khi \(x=y=z;x,y,z>0\)
Theo bất đẳng thức tam giác
\(\Rightarrow\left\{\begin{matrix}a< b+c\\b< c+a\\c< a+b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}b+c-a>0\\c+a-b>0\\a+b-c>0\end{matrix}\right.\)
Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)
\(\Rightarrow\left\{\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{a+c-b}\ge\dfrac{2}{a}\end{matrix}\right.\)
Cộng theo từng vế
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( đpcm )
Nội suy Sửa đề làm cho bạn
Bài 1:
\(a^2+b^2+c^2\ge ab+bc+ac+\dfrac{\left(a-b\right)^2}{26}+\dfrac{\left(b-c\right)^2}{2}+\dfrac{\left(c-a\right)^2}{2009}\)Nhân 2 chuyển Vế
\(2a^2+2b^2+2c^2-2ab-2bc-2ac-\left[\dfrac{\left(a-b\right)^2}{13}+\dfrac{\left(b-c\right)^2}{3}+\dfrac{2\left(c-a\right)^2}{2009}\right]\ge0\)Ghép Bình phướng
\(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2-\left[\dfrac{\left(a-b\right)^2}{13}+\dfrac{\left(b-c\right)^2}{3}+\dfrac{2.\left(c-a\right)^2}{2009}\right]\ge0\)Ghép nhân tử
\(\left[\left(a-b\right)^2\left(1-\dfrac{1}{13}\right)+\left(b-c\right)^2\left(1-\dfrac{1}{3}\right)+\left(c-a\right)^2\left(1-\dfrac{2}{2009}\right)\right]\ge0\)
Thu gọn có thể không cần
\(\left[\left(a-b\right)^2\left(\dfrac{12}{13}\right)+\left(b-c\right)^2\left(\dfrac{2}{3}\right)+\left(c-a\right)^2\left(\dfrac{207}{2009}\right)\right]\ge0\)VT là tổng 3 số không âm
Đẳng thức khi a=b=c
=> dpcm
1) 2( a2 + b2 ) ≥ ( a + b)2
<=> 2a2 + 2b2 - a2 - 2ab - b2 ≥ 0
<=> a2 - 2ab + b2 ≥ 0
<=> ( a - b )2 ≥ 0 ( luôn đúng )
=> đpcm
2) Áp dụng BĐT Cô-si cho 2 số dương x , y , ta có :
a + b ≥ \(2\sqrt{ab}\)
=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ 2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)
=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\) ) ≥ \(2\sqrt{xy}\)2\(\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}\)
=> ( x + y)( \(\dfrac{1}{x}+\dfrac{1}{y}\)) ≥ 4
=> \(\dfrac{1}{x}+\dfrac{1}{y}\) ≥ \(\dfrac{4}{x+y}\)
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 )
a)
\(a^2+b^2+c^2+d^2+m^2-a(b+c+d+m)\)
\(=\frac{4a^2+4b^2+4c^2+4d^2+4m^2-4a(b+c+d+m)}{4}\)
\(=\frac{(a^2+4b^2-4ab)+(a^2+4c^2-4ac)+(a^2+4d^2-4ad)+(a^2+4m^2-4am)}{4}\)
\(=\frac{(a-2b)^2+(a-2c)^2+(a-2d)^2+(a-2m)^2}{4}\geq 0\) (đpcm)
Dấu "=" xảy ra khi \(a=2b=2c=2d=2m\)
b)
Xét hiệu
\(\frac{1}{x}+\frac{1}{y}-\frac{4}{x+y}=\frac{x+y}{xy}-\frac{4}{x+y}=\frac{(x+y)^2-4xy}{xy(x+y)}\)
\(=\frac{x^2+y^2-2xy}{xy(x+y)}=\frac{(x-y)^2}{xy(x+y)}\geq 0, \forall x,y>0\)
\(\Rightarrow \frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\) (đpcm)
Dấu "=" xảy ra khi $x=y$
c)
Xét hiệu:
\((a^2+c^2)(b^2+d^2)-(ab+cd)^2\)
\(=(a^2b^2+a^2d^2+c^2b^2+c^2d^2)-(a^2b^2+2abcd+c^2d^2)\)
\(=a^2d^2-2abcd+b^2c^2=(ad-bc)^2\geq 0\)
\(\Rightarrow (a^2+c^2)(b^2+d^2)\geq (ab+cd)^2\) (đpcm)
Dấu "=" xảy ra khi \(ad=bc\)
d)
Xét hiệu:
\(a^2+b^2-(a+b-\frac{1}{2})=a^2+b^2-a-b+\frac{1}{2}\)
\(=(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})\)
\(=(a-\frac{1}{2})^2+(b-\frac{1}{2})^2\geq 0\)
\(\Rightarrow a^2+b^2\geq a+b-\frac{1}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{1}{1-ab}=1+\dfrac{ab}{1-ab}\le1+\dfrac{ab}{1-\dfrac{a^2+b^2}{2}}=1+\dfrac{2ab}{a^2+b^2+2c^2}\)
\(=1+\dfrac{2ab}{\left(a^2+c^2\right)+\left(b^2+c^2\right)}\le1+\dfrac{ab}{\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}}\)
\(\le1+\dfrac{1}{2}\left(\dfrac{a^2}{a^2+c^2}+\dfrac{b^2}{b^2+c^2}\right)\). Tương tự ta cũng có:
\(\dfrac{1}{1-bc}\le1+\dfrac{1}{2}\left(\dfrac{b^2}{a^2+b^2}+\dfrac{c^2}{a^2+c^2}\right);\dfrac{1}{1-ca}\le1+\dfrac{1}{2}\left(\dfrac{c^2}{b^2+c^2}+\dfrac{a^2}{a^2+b^2}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le3+\dfrac{1}{2}\left(\dfrac{a^2+b^2}{a^2+b^2}+\dfrac{b^2+c^2}{b^2+c^2}+\dfrac{c^2+a^2}{c^2+a^2}\right)=\dfrac{9}{2}\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
2)
Xét hiệu:
\(A^2+B^2+C^2+D^2+4-2A-2B-2C-2D\)
\(=\left(A^2-2A+1\right)+\left(B^2-2B+1\right)+\left(C^2-2C+1\right)+\left(D^2-2D+1\right)\)
\(=\left(A-1\right)^2+\left(B-1\right)^2+\left(C-1\right)^2+\left(D-1\right)^2\ge0\)
=> BĐT luôn đúng
Vậy \(A^2+B^2+C^2+D^2+4\ge2\left(A+B+C+D\right)\)
1)
Áp dụng BĐT Cauchy cho 2 số không âm, ta có:
\(\dfrac{AB}{C}+\dfrac{BC}{A}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{BC}{A}}=2B\) (1)
\(\dfrac{BC}{A}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{BC}{A}.\dfrac{AC}{B}}=2C\) (2)
\(\dfrac{AB}{C}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{AC}{B}}=2A\) (3)
Từ (1)(2)(3) cộng vế theo vế:
\(2\left(\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\right)\ge2\left(A+B+C\right)\)
\(\Rightarrow\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\ge A+B+C\)