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a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
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.
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 \(x=y=z;x,y,z>0\)
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 )
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:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
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 \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Ta có: \(\left\{{}\begin{matrix}abc=1\\\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3b^3c^3=1\\\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\end{matrix}\right.\)
\(a^3b^3+b^3c^3+c^3a^3=a^3b^3c^3\left(\dfrac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}\right)=1.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
\(\Rightarrow S=\left(a^3b^3+b^3c^3+c^3a^3\right)\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2\)
Lại có:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2-\dfrac{1}{c}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c^2}\right)-\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{-3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{-3}{ab}\left(\dfrac{-1}{c}\right)=\dfrac{3}{abc}=3\)
\(\Rightarrow S=\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)^2=3^2=9\)