Help me phần b ,
a, CMR : \(\dfrac{a^2}{x}\) + \(\dfrac{b^2}{y}\)\(\ge\)\(\dfrac{\left(a+b\right)^2}{x+y}\)
b, CMR : \(\dfrac{1}{a^2+2bc}\)+ \(\dfrac{1}{b^2+2ac}\) + \(\dfrac{1}{c^2+2bc}\)\(\ge\) 9
Những câu hỏi liên quan
Cho ba số a,b,c dương abc=1. CMR
P = \(\dfrac{a^2}{\left(2ab+1\right)\left(ab+2\right)}+\dfrac{b^2}{\left(2bc+1\right)\left(bc+2\right)}+\dfrac{c^2}{\left(2ac+1\right)\left(ac+2\right)}\)\(\ge\)\(\dfrac{1}{3}\)
Cho a, b, c > 0 thỏa mãn abc = 1.
CMR : \(\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2ab+2bc+2ac}\ge\dfrac{9}{2}\)
\(VT\ge a+b+c+\dfrac{9}{2\left(ab+bc+ca\right)}\ge\sqrt{3\left(ab+bc+ca\right)}+\dfrac{9}{2\left(ab+bc+ca\right)}\)
\(=\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{\sqrt{3\left(ab+bc+ca\right)}}{2}+\dfrac{9}{2\left(ab+bc+ca\right)}\ge3\sqrt[3]{\dfrac{27}{8}}=\dfrac{9}{2}\)
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Áp dụng BĐT Cauchy ta có
\(\dfrac{b^2}{a}+a\ge2b;\) \(\dfrac{c^2}{b}+b\ge2c\); \(\dfrac{a^2}{c}+c\ge2a\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}\ge a+b+c\)
\(\Rightarrow\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2\left(ab+bc+ac\right)}\ge a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\)Ta phải chứng minh
\(a+b+c+\dfrac{9}{2\left(ab+bc+ac\right)}\ge\dfrac{9}{2}\)
\(\Leftrightarrow4\left(a+b+c\right)\left(ab+bc+ac\right)+18\ge18\left(ab+bc+ac\right)\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge0\)
Áp dụng BĐT Cauchy:
\(ab+bc+ac\ge3\sqrt[3]{a^2b^2c^2}=3\)
\(a+b+c\ge3\sqrt[3]{abc}=3\)
\(\Rightarrow\left(ab+bc+ac\right)\left(4\left(a+b+c\right)-18\right)+18\ge3\left(4.3-18\right)+18=0\)=> đpcm
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Cho a, b, c > 0 thỏa mãn abc = 1.
CMR : \(\dfrac{b^2}{a}+\dfrac{c^2}{b}+\dfrac{a^2}{c}+\dfrac{9}{2ab+2bc+2ac}\ge\dfrac{9}{2}\)
Cho ba số a, b, c thỏa mãn điều kiện: \(a^2+b^2+c^2=1\).
CMR: \(\dfrac{a^2}{1+2bc}+\dfrac{b^2}{1+2ac}+\dfrac{c^2}{1+ab}\ge\dfrac{3}{5}\)
3 tháng 4 2022 lúc 20:53
B1: Cho 0le a,b,cle2 thỏa mãn a+b+c3. CMR: a^2+b^2+c^2le5B2: Cho a,bge0 thỏa mãn a^2+b^2a+b. TÌm GTLN Sdfrac{a}{a+1}+dfrac{b}{b+1}B3: CMR: dfrac{1}{left(x-yright)^2}+dfrac{1}{x^2}+dfrac{1}{y^2}gedfrac{4}{xy}forall xne y,xyne0
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B1: Cho \(0\le a,b,c\le2\) thỏa mãn \(a+b+c=3\). CMR: \(a^2+b^2+c^2\le5\)
B2: Cho \(a,b\ge0\) thỏa mãn \(a^2+b^2=a+b\). TÌm GTLN \(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}\)
B3: CMR: \(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
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5 tháng 5 2022 lúc 22:06
1. Cho a,b,c t/m: \(\left\{{}\begin{matrix}a\ge\dfrac{4}{3}\\b\ge\dfrac{4}{3}\\c\ge\dfrac{4}{3}\end{matrix}\right.\) và \(a+b+c=6\)
\(CMR:\dfrac{a}{a^2+1}+\dfrac{b}{b^2+1}+\dfrac{c}{c^2+1}\ge\dfrac{6}{5}\)
2. Cho x,y >0 t/m: \(2x+3y-13\ge0\)
Tìm min \(P=x^2+3x+\dfrac{4}{x}+y^2+\dfrac{9}{y}\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
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Giúp mk bài toán nang cao này nhé mm
a) Cho x,y,z là số dương CMR
\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)lớn hơn hoặc bằng 9
b) a,b,c > 0 thoả mãn a+b+c bé hơn học bằng 1 CMR
\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2bc}+\dfrac{1}{c^2+2bc}\)lớ hơn hoặc bằng 9
_________Thanks na_______
Câu a :
Theo BĐT cauchy schwar ta có :
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}\)
\(\Rightarrow\left(x+y+z\right)\left(\dfrac{9}{x+y+z}\right)\ge9\)
Câu b : Sửa lại đề nha :
Theo BĐT cauchy schwar ta có :
\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\dfrac{9}{\left(a+b+c\right)^2}\)
Vì \(a+b+c\le\Rightarrow\left(a+b+c\right)^2\le1\)
\(\Rightarrow\) \(\dfrac{9}{\left(a+b+c\right)^2}\ge9\)
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7 tháng 2 2021 lúc 20:26
cho a,b,c>0 và \(a+b+c\le1\)
cmr \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge9\)
Bài 1: \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\)
CMR:\(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\)
Bài 2:Cho x=\(\dfrac{b^2+c^2-a^2}{2bc}\);y=\(\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)
Tính A=x+y+xy
Mấy bài này đăng nhiều rồi bạn ;v
Bài 1: Nhân cả 2 vế cho a+b+c rồi rút gọn được đpcm
Bài 2: Thêm 1 rồi bớt 1 :v (x+y+xy+1-1)
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