ai giải giùm mình với ạ
cho \(a^2+b^2+c^2=3\) và a,b,c >0
CMR \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{9}{a+b+c}\)
Những câu hỏi liên quan
1
cho a,b,c>0: abc=1
cm:\(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+\frac{9}{2\left(ab+bc+ca\right)}\ge\frac{9}{2}\)
2)
cho a,b,c >0 thỏa a+b+c=3. cm:
\(\frac{3+a^2}{b+c}+\frac{3+b^2}{c+a}+\frac{3+c^2}{a+b}\ge6\)
làm đc bài nào thì trình bàybài giải vào giùm mik lun nhé!
thanks
2) \(VT=\left(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\right)+3\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
Xét \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\) (1)
Xét \(3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
Áp dụng bất đẳng thức Cauchy dạng phân thức
\(\Rightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)
\(\Rightarrow3\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge3.\frac{3}{2}=\frac{9}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\frac{9}{2}+\frac{3}{2}=6\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
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cho a,b,c >0. CMR:
+) \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}\)
+)\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge ab+bc+ca\)
Giups mình giải mấy câu trên với
cho a,b,c >0 chứng minh \(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\) ai làm dùm em với ạ
Cho a,b,c là các số thực dương. CMR: \(\frac{a}{c^2}+\frac{b}{a^2}+\frac{c}{b^2}\ge\frac{9}{a+b+c}\)
Ai nhanh và đúng thì mình sẽ tick và add friends nhé. Thanks. Please help me!!! PLEASE!!!
\(VT-VP=\Sigma_{cyc}\frac{2a+b+c}{a^2b\left(a+b+c\right)}\left(a-b\right)^2\ge0\)
hay \(\frac{a}{c^2}+\frac{1}{a}\ge\frac{2}{c}\)\(\Leftrightarrow\)\(\frac{a}{c^2}\ge\frac{2}{c}-\frac{1}{a}\)\(\Rightarrow\)\(VT\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
"=" \(\Leftrightarrow\)\(a=b=c\)
C/m các BĐT sau :
1.a^3-3a+4ge b^3-3b
2,frac{1}{frac{1}{a+c}+frac{1}{b+d}}gefrac{1}{frac{1}{a}+frac{1}{b}}+frac{1}{frac{1}{c}+frac{1}{d}} với a, b, c, d0
3,a^3+b^3gefrac{1}{4};a+bge1
4, a^3+b^3le a^4+b^4;a+bge2
5, left(a+bright)left(a^3+b^3right)left(a^5+b^5right)le4left(a^9+b^9right);a,bge0
6, frac{c+a}{sqrt{a^2+c^2}}gefrac{c+b}{sqrt...
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C/m các BĐT sau :
\(1.a^3-3a+4\ge b^3-3b
\)
\(2,\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}}\ge\frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{c}+\frac{1}{d}}\) với a, b, c, d>0
\(3,a^3+b^3\ge\frac{1}{4};a+b\ge1\)
4, \(a^3+b^3\le a^4+b^4;a+b\ge2\)
5, \(\left(a+b\right)\left(a^3+b^3\right)\left(a^5+b^5\right)\le4\left(a^9+b^9\right);a,b\ge0\)
6, \(\frac{c+a}{\sqrt{a^2+c^2}}\ge\frac{c+b}{\sqrt{c^2+b^2}};a>b>0,c>\sqrt{ab}\)
Các bn làm đc bài nào thì giúp mk với, cảm ơn ạ !
1.
C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)
\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)
\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)
\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)
\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)
Theo bài ra: \(a^3-b^3\ge3a-3b-4\)
\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)
Thật vậy:
\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)
\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))
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2.
\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)
\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)
\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)
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Cho a, b, c > 0. CMR:
a, \(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge2\left(a+b+c\right)\)
b, \(\frac{a^2}{b^3}+\frac{b^2}{c^3}+\frac{c^2}{a^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
c, \(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
Giúp mình với các bạn ơiii
a) Bổ đề: \(x^3+y^3\ge xy\left(x+y\right)\forall x,y>0\)
\(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge\frac{ab\left(a+b\right)}{ab}+\frac{bc\left(b+c\right)}{bc}+\frac{ca\left(c+a\right)}{ca}=2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
Cảm ơn bạn nhiều nhé Nhật Pháp soi chiếu thế gian. Nếu có thể, mong bạn hãy giúp mình những phần còn lại ^^
c) Áp dụng bất đẳng thức AM-GM:
\(\frac{a^3}{b}+\frac{a^3}{b}+b^2\ge3\sqrt[3]{\frac{a^3}{b}.\frac{a^3}{b}.b^2}=3a^2\);
\(\frac{b^3}{c}+\frac{b^3}{c}+c^2\ge3\sqrt[3]{\frac{b^3}{c}.\frac{b^3}{c}.c^2}=3b^2\);
\(\frac{c^3}{a}+\frac{c^3}{a}+a^2\ge3\sqrt[3]{\frac{c^3}{a}.\frac{c^3}{a}.a^2}=3c^2\)
Cộng theo từng vế ba bất đẳng thức trên ta đươc:
\(2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+a^2+b^2+c^2\ge3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge a^2+b^2+c^2\)
Dấu "=" xảy ra khi \(a=b=c\)
Xem thêm câu trả lời
\(\frac{6+a}{9-a^2}+...=\frac{6}{9-a^2}+...+\frac{a^2}{9a-a^3}\ge\frac{54}{27-a^2-b^2-c^2}+\frac{\left(a+b+c\right)^2}{9\left(a+b+c\right)-\left(a^3+b^3+c^3\right)}\)
\(\ge\frac{54}{27-2\left(a+b+c\right)+3}+\frac{9}{27-3\left(a+b+c\right)+6}=\frac{54}{24}+\frac{9}{24}=\frac{21}{8}\)
đây là toán đâu phải văn. bạn bị say rượu à
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áp dụng cô si ta có:+)frac{a^5}{b^3}+frac{a^3}{b}gefrac{2a^4}{b^2};frac{b^5}{c^3}+frac{b^3}{c}gefrac{2b^4}{c^2};frac{c^5}{a^3}+frac{c^3}{a}gefrac{2c^4}{a^2}Leftrightarrowfrac{a^5}{b^3}+frac{b^5}{c^3}+frac{c^5}{a^3}ge2left(frac{a^4}{b^2}+frac{b^4}{c^2}+frac{c^4}{a^2}right)-left(frac{a^3}{b}+frac{b^3}{c}+frac{c^3}{a}right)+)frac{a^4}{b^2}+a^2gefrac{2a^3}{b};frac{b^4}{c^2}+b^2gefrac{2b^3}{c};frac{c^4}{a^2}+c^2gefrac{2C^3}{a}Leftrightarrowfrac{a^4}{b^2}+frac{b^4}{c^2}+frac{c^4}{a^2}ge2left(frac{a^3}...
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áp dụng cô si ta có:
+)\(\frac{a^5}{b^3}+\frac{a^3}{b}\ge\frac{2a^4}{b^2};\frac{b^5}{c^3}+\frac{b^3}{c}\ge\frac{2b^4}{c^2};\frac{c^5}{a^3}+\frac{c^3}{a}\ge\frac{2c^4}{a^2}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge2\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)-\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)\)
+)\(\frac{a^4}{b^2}+a^2\ge\frac{2a^3}{b};\frac{b^4}{c^2}+b^2\ge\frac{2b^3}{c};\frac{c^4}{a^2}+c^2\ge\frac{2C^3}{a}\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge2\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)-\left(a^2+b^2+c^2\right)\)
+)\(\frac{a^3}{b}+ab\ge2a^2;\frac{b^3}{c}+bc\ge2b^2;\frac{c^3}{a}+ca\ge2c^2\)
\(\Leftrightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\left(a^2+b^2+c^2\right)+\left(a^2+b^2+c^2-ab-bc-ca\right)\ge\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\ge\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)+\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}-a^2-b^2-c^2\right)\ge\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\)
\(\Leftrightarrow\frac{a^5}{b^3}+\frac{b^5}{c^3}+\frac{c^5}{a^3}\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)+\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}-\frac{a^3}{b}-\frac{b^3}{c}-\frac{c^3}{a}\right)\ge\left(\frac{a^4}{b^2}+\frac{b^4}{c^2}+\frac{c^4}{a^2}\right)\)
Cho a,b,c >0 cmr
a, \(\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\ge\frac{9}{a+b+c}\)
b, \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)