CMR mọi a,b,c thuộc R ta có a) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
b) \(a^2+b^2+c^2\ge ab+bc+ca\)
c) \(a^4+b^4\ge a^3b+ab^3\)
Những câu hỏi liên quan
Bài 1 :Cho a,b,c dương thỏa mãn a+b+c=2
CMR \(\frac{bc}{\sqrt{3a^2+4}}+\frac{ca}{\sqrt{3b^2+4}}+\frac{ab}{\sqrt{3c^2+4}}\ge\frac{\sqrt{3}}{3}\)
Bài 2:Cho a,b,c>0. CMR
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ca\right)\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)
\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)
\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)
\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)
\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)
\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)
Từ ( 1 ) và ( 2 ) có đpcm
Cho a,b,c0CMR :frac{a^4}{bleft(b+cright)}+frac{b^4}{cleft(c+aright)}+frac{c^4}{aleft(a+bright)}gefrac{1}{2}left(ab+bc+caright)Áp dụng bđt Svac-xo ta có :VTgefrac{left(a^2+b^2+c^2right)^2}{a^2+b^2+c^2+ab+bc+ca}gefrac{left(a^2+b^2+c^2right)^2}{2left(a^2+b^2+c^2right)}frac{a^2+b^2+c^2}{2}gefrac{ab+bc+ca}{2}Dấu - xảy ra abc
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Cho \(a,b,c>0\)
CMR :\(\frac{a^4}{b\left(b+c\right)}+\frac{b^4}{c\left(c+a\right)}+\frac{c^4}{a\left(a+b\right)}\ge\frac{1}{2}\left(ab+bc+ca\right)\)
Áp dụng bđt Svac-xo ta có :
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+ab+bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2}{2}\ge\frac{ab+bc+ca}{2}\)
Dấu "-" xảy ra \(< =>a=b=c\)
Cho a,b,c∈R.CM bđt a^2+b^2+c^2ge ab+bc+ca (1). Áp dụng cm các bđt sau:
a)left(a+b+cright)^2le3left(a^2+b^2+c^2right)
b)frac{a^2+b^2+c^2}{3}geleft(frac{a+b+c}{3}right)^2
c)left(a+b+cright)^2ge3left(ab+bc+caright)
d)a^4+b^4+c^4ge abcleft(a+b+cright)
e)frac{a+b+c}{3}gesqrt{frac{ab+bc+ca}{3}}vớia,b,c0
f)a^4+b^4+c^4ge abc nếu a+b+c1
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Cho a,b,c∈R.CM bđt \(a^2+b^2+c^2\ge ab+bc+ca\) (1). Áp dụng cm các bđt sau:
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
b)\(\frac{a^2+b^2+c^2}{3}\ge\left(\frac{a+b+c}{3}\right)^2\)
c)\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d)\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
e)\(\frac{a+b+c}{3}\ge\sqrt{\frac{ab+bc+ca}{3}}vớia,b,c>0\)
f)\(a^4+b^4+c^4\ge abc\) nếu a+b+c=1
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) (luôn đúng)
a/ Từ BĐT ban đầu ta có:
\(2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) (đpcm)
b/ Chia 2 vế của BĐT ở câu a cho 9 ta được:
\(\frac{a^2+b^2+c^2}{3}\ge\frac{\left(a+b+c\right)^2}{9}=\left(\frac{a+b+c}{3}\right)^2\) (đpcm)
c/ Cộng 2 vế của BĐT ban đầu với \(2ab+2bc+2ca\) ta được:
\(a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d/ Áp dụng BĐT ban đầu cho các số \(a^2;b^2;c^2\) ta được:
\(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\ge a^2b^2+b^2c^2+c^2a^2\)
Mặt khác ta cũng có:
\(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge ab.bc+bc.ca+ab+ca=abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
e/ Chia 2 vế của BĐT ở câu c cho 9 ta được:
\(\frac{\left(a+b+c\right)^2}{9}\ge\frac{ab+bc+ca}{3}\)
Khai căn 2 vế: \(\Rightarrow\frac{a+b+c}{3}\ge\sqrt{\frac{ab+bc+ca}{3}}\)
f/ Áp dụng BĐT ở câu d:
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)=abc\) (do \(a+b+c=1\))
CM BĐT:
a) \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
b) \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\)
c) \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
d) \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
a)\(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Leftrightarrow3a^2+3b^2+3c^2-a^2-b^2-c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng)
b,c tương tự
d)Áp dụng bđt AM-GM ta được
\(a^4+a^4+b^4+c^4\ge4\sqrt[4]{a^4a^4b^4c^4}=4a^2bc\)
TT\(\Rightarrow a^4+b^4+b^4+c^4\ge4ab^2c\)
\(a^4+b^4+c^4+c^4\ge4abc^2\)
Cộng vế theo vế ta được \(4\left(a^4+b^4+c^4\right)\ge4\left(a^2bc+ab^2c+abc^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\left(đpcm\right)\)
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d)
\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-ab^2c-abc^2\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2ab^2c-2abc^2\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2+2a^2c^2-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)+\left(b^2c^2+c^2a^2-2c^2abc\right)+\left(a^2b^2+c^2a^2-2a^2ab\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ac\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a , b , c
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ta có:3left(a^4+b^4+c^4right)geleft(a^2+b^2+c^2right)^2geleft(ab^2+bc^2+ca^2right)left(a+b+cright)Rightarrow a^4+b^4+c^4ge a+b+clại có:left(a^4+b^4+c^4right)left(b^2+a^2+c^2right)geleft(ab^2+bc^2+ca^2right)^29Rightarrowleft(a^4+b^4+c^4right).sqrt{3left(a^4+b^4+c^4right)}ge9Rightarrow a^4+b^4+c^4ge3Rightarrow24left(a^4+b^4+c^4right)ge a+b+c+69ge12sqrt[3]{a+7}+...
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ta có:
\(3\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\ge\left(ab^2+bc^2+ca^2\right)\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4\ge a+b+c\)
lại có:
\(\left(a^4+b^4+c^4\right)\left(b^2+a^2+c^2\right)\ge\left(ab^2+bc^2+ca^2\right)^2=9\)
\(\Rightarrow\left(a^4+b^4+c^4\right).\sqrt{3\left(a^4+b^4+c^4\right)}\ge9\)
\(\Rightarrow a^4+b^4+c^4\ge3\)
\(\Rightarrow24\left(a^4+b^4+c^4\right)\ge a+b+c+69\ge12\sqrt[3]{a+7}+...\)
cho a,b,c là số thực dương. Cmr:
1.\(\dfrac{a}{b^2+bc+c^2}+\dfrac{b}{c^2+ca+a^2}+\dfrac{c}{a^2+ab+b^2}\ge\dfrac{a+b+c}{ab+bc+ca}\)
2.\(\left(a+b+c\right)\left(\dfrac{a}{\left(b+c\right)^2}+\dfrac{b}{\left(c+a\right)^2}+\dfrac{c}{\left(a+b\right)^2}\right)\ge\dfrac{9}{4}\)
Bài 1
\(VT=\dfrac{a^2}{ab^2+abc+ac^2}+\dfrac{b^2}{c^2b+abc+a^2b}+\dfrac{c^2}{a^2c+abc+b^2c}\)
Áp dụng bđt Cauchy dạng phân thức
\(\Rightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)
\(\Leftrightarrow VT\ge\dfrac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)
\(\Leftrightarrow VT\ge\dfrac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)
Dấu ''='' xảy ra khi \(a=b=c\)
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Bài 2
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\dfrac{\sqrt{a}}{b+c}\right)^2+\left(\dfrac{\sqrt{b}}{c+a}\right)^2+\left(\dfrac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dụng bđt Bunhiacopxki ta có
\(VT\ge\left(\sqrt{a}.\dfrac{\sqrt{a}}{b+c}+\sqrt{b}.\dfrac{\sqrt{b}}{c+a}+\sqrt{c}.\dfrac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Áp dụng bđt Cauchy dạng phân thức ta có
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)^2\ge\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)
\(\Rightarrow VT\ge\dfrac{9}{4}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c\)
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CMR:
a/\(a^2+b^2+c^2\ge\text{ab}+bc+c\text{a}\)
b/\(3\left(\text{a}b+bc+c\text{a}\right)\le\left(\text{a}+b+c\right)^2\le3\left(\text{a}^2+b^2+c^2\right)\)
c/\(\text{a}^3+b^3\ge\text{a}b\left(\text{a}+b\right)\)
a) \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b=c\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
+) vế 1 bđt \(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
+) vế 2 bđt \(\Leftrightarrow3a^2+3b^2+3c^2\ge a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)( CMTT câu a )
Từ đây ta có đpcm
Dấu "=" \(\Leftrightarrow a=b=c\)
c) \(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)
\(\Leftrightarrow a^2-ab+b^2\ge ab\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\)( luôn đúng )
Dấu "=" \(\Leftrightarrow a=b\)
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Chứng minh rằng với mọi số thực a, b ta luôn có:
a)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
b)\(a^2+b^2+c^2\ge ab+bc+ca\)
`a) 2 ( a^2 + b^2 ) >= ( a + b )^2`
`<=> 2a^2 + 2b^2 >= a^2 + 2ab + b^2`
`<=> a^2 - 2ab + b^2 >= 0`
`<=> ( a - b )^2 >= 0` (Luôn đúng `AA a,b`)
`=>` Đẳng thức được c/m
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`b) a^2 + b^2 + c^2 >= ab + bc + ca`
`<=> 2a^2 + 2b^2 + 2c^2 >= 2ab + 2bc + 2ca`
`<=> ( a^2 - 2ab + b^2 ) + ( b^2 - 2bc + c^2 ) + ( c^2 - 2ca + a^2 ) >= 0`
`<=> ( a - b )^2 + ( b - c )^2 + ( c - a )^2 >= 0` (Luôn đúng `AA a,b,c`)
`=>` Đẳng thức được c/m
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Cho a,b,c > 0. CMR: (a + b + c)2 \(\ge\) 3(ab + bc + ca)
và \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}+\frac{ab+bc+ca}{\left(a+b+c\right)^2}\ge\frac{10}{3}\)
