cho a,b, c > hoac = 0 va a+b+c=1.chung minh
\(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}>3.5\)
2 cho a,b,c >0 . chung minh
\(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}>hoac=3\)
Những câu hỏi liên quan
1.Chứng minh sqrt{x^2+xy+y^2}+sqrt{x^2+xz+z^2}gesqrt{y^2+yz+z^2}2. Cho a,b,c0. Chứng minh left(sqrt[3]{a}+sqrt[3]{b}+sqrt[3]{c}right)left(frac{1}{sqrt[3]{a}}+frac{1}{sqrt[3]{b}}+frac{1}{sqrt[3]{c}}right)-frac{a+b+c}{sqrt[3]{abc}}le63. Cho a,b0 , n là số nguyên dương. Chứng minh frac{1}{sqrt[n]{a}}+frac{1}{sqrt[n]{b}}ge2sqrt[n]{frac{2}{a+b}}4. Cho a,b,c 0. Chứng minh frac{1}{a^2+bc}+frac{1}{b^2+ca}+frac{1}{c^2+ba}lefrac{a+b+c}{2abc}
Đọc tiếp
1.Chứng minh \(\sqrt{x^2+xy+y^2}+\sqrt{x^2+xz+z^2}\ge\sqrt{y^2+yz+z^2}\)
2. Cho a,b,c>0. Chứng minh \(\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}+\frac{1}{\sqrt[3]{c}}\right)-\frac{a+b+c}{\sqrt[3]{abc}}\le6\)
3. Cho a,b>0 , n là số nguyên dương. Chứng minh \(\frac{1}{\sqrt[n]{a}}+\frac{1}{\sqrt[n]{b}}\ge2\sqrt[n]{\frac{2}{a+b}}\)
4. Cho a,b,c >0. Chứng minh \(\frac{1}{a^2+bc}+\frac{1}{b^2+ca}+\frac{1}{c^2+ba}\le\frac{a+b+c}{2abc}\)
Cho a+b+c=0 va a,b,c≠0. Chứng minh đẳng thức:
\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
Ta có
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=\frac{a^2bc+ab^2c+abc^2}{a^2b^2c^2}=\frac{abc\left(a+b+c\right)}{a^2b^2c^2}=0\)
Ta lại có
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
Từ đó
\(\Rightarrow\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)
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Bài 1: Cho a,b>0. Chứng minh \(\sqrt[3]{\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)}< \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\)
Bài 2: Cho a,b>0. Chứng minh \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge\frac{2\sqrt{2}}{\sqrt{a+b}}\)
Bài 3: Cho a,b,c>0. Chứng minh \(\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\le a+b+c\)
cho so thuc a,b,c voi a ,b duong va c\(\ne\)0 thoa man
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
1/chung minh c<0 , a+c>0 va b+c >0
2/chung minh \(\sqrt{a+b}=\sqrt{a+c}+\sqrt{b+c}\)
cho a,b,c>0 và a+b+c>=6
chứng minh S=\(\sqrt{a^2+\frac{1}{b+c}}+\sqrt{b^2+\frac{1}{c+a}}+\sqrt{c^2+\frac{1}{a+b}}>=\frac{3\sqrt{17}}{2}\)
Áp dụng bất đẳng thức Min.cop.xki
\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)
Dấu "=" xảy ra khi \(\frac{a}{c}=\frac{b}{d}\) (Chứng minh bằng biến đổi tương đương)
Áp dụng:
\(S=\sqrt{a^2+\frac{1}{b+c}}+\sqrt{b^2+\frac{1}{c+a}}+\sqrt{c^2+\frac{1}{a+b}}\ge\sqrt{\left(a+b\right)^2+\left(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\right)^2}+\sqrt{c^2+\frac{1}{a+b}}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{c+a}}\right)^2}\)
\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)^2}\)
Theo Bunhiacopxki: \(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left(a+b+b+c+c+a\right)=6\left(a+b+c\right)\)
\(\Rightarrow\left(a+b+c\right)^2+\frac{81}{\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2}\ge\left(a+b+c\right)^2+\frac{81}{6\left(a+b+c\right)}\)
\(=\frac{\left(a+b+c\right)^2}{32}+\frac{81}{12\left(a+b+c\right)}+\frac{81}{12\left(a+b+c\right)}+\frac{31}{32}\left(a+b+c\right)^2\)
\(\ge3\sqrt[3]{\frac{\left(a+b+c\right)^2}{32}.\frac{81}{12\left(a+b+c\right)}.\frac{81}{12\left(a+b+c\right)}}+\frac{31}{32}.6^2\)
\(=\frac{153}{4}=\left(\frac{3\sqrt{17}}{2}\right)^2\)
\(\Rightarrow S\ge\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=2\).
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Cho a,b,c > 0 . Chứng minh rằng: \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge2\left(\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\right)\)
Theo BĐT AM-GM :
\(\sqrt{b}=\sqrt{b\cdot1}\le\frac{b+1}{2}\)
\(\Rightarrow\frac{a}{\sqrt{b}}\ge\frac{a}{\frac{b+1}{2}}=\frac{2a}{b+1}\)
Dấu "=" xảy ra \(\Leftrightarrow b=1\)
+ Tương tự ta cm đc :
\(\frac{b}{\sqrt{c}}\ge\frac{2b}{c+1}\). Dấu "=" xảy ra \(\Leftrightarrow c=1\)
\(\frac{c}{\sqrt{a}}\ge\frac{2c}{a+1}\). Dấu "=" xảy ra \(\Leftrightarrow a=1\)
Do đó : \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\ge2\left(\frac{a}{b+1}+\frac{b}{c+}+\frac{c}{a+1}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
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chứng minh \(\frac{a}{1+a}+\frac{2b}{2+b}+\frac{3c}{3+c}< hoac=\frac{6}{7}\)cho a,b,c>0.a+b+c=1
ta có:\(\frac{a}{a+1}=1-\frac{1}{a+1};\frac{2b}{2+b}=2-\frac{4}{2+b};\frac{3c}{3+c}=3-\frac{9}{3+c}\)
\(\Rightarrow\frac{a}{1+a}+\frac{2b}{2+b}+\frac{3c}{3+c}\le\left(1+2+3\right)-\left(\frac{1}{a+1}+\frac{4}{b+2}+\frac{9}{c+3}\right)\)
\(\le6-\frac{\left(1+2+3\right)^2}{a+b+c+1+2+3}=6-\frac{36}{7}=\frac{6}{7}\left(Q.E.D\right)\)
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Cho a, b, c > 0 thoả mãn: a + b + c = \(\sqrt{a}+\sqrt{b}+\sqrt{c}\) = 2. Chứng minh: \(\frac{\sqrt{a}}{a+1}+\frac{\sqrt{b}}{b+1}+\frac{\sqrt{c}}{c+1}=\frac{2}{\sqrt{\left(1+a\right)\left(1+b\left(1+c\right)\right)}}\)
Cho a, b, c > 0 thoả mãn a + b + c = \(\sqrt{a}+\sqrt{b}+\sqrt{c}\) = 2. Chứng minh: \(\frac{\sqrt{a}}{a+1}+\frac{\sqrt{b}}{b+1}+\frac{\sqrt{c}}{c+1}=\frac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
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Câu hỏi của Nguyễn Thu Trà - Toán lớp 9 | Học trực tuyến
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