Cho a,b,c>0.Chứng minh rằng\(\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le1\)
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Cho a, b, c dương. CMR:
\(\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le1\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{a+\sqrt{\left(\sqrt{ac}+\sqrt{ab}\right)^2}}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{c}+\sqrt{b}}\)
Tương tự cho 2 BĐT còn lại rồi cộng theo vế:
\(VT\le\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1=VP\)
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Cho a, b, c dương. CMR:
\(\dfrac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{b+\sqrt{\left(b+c\right)\left(b+a\right)}}+\dfrac{c}{c+\sqrt{\left(c+a\right)\left(c+b\right)}}\le1\)
Áp dụng bđt Bu-nhi-a, ta có
\(\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ab}+\sqrt{ac}\)
=>\(\frac{a}{a+\sqrt{\left(a+b\right)\left(a+c\right)}}\le\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
Tương tự, rồi + vào, ta có
A\(\le\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\) (ĐPCM)
dấu =xảy ra <=>a=b=c>o
^_^
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từ giả thiết, ta có dfrac{1}{xy}+dfrac{1}{yz}+dfrac{1}{zx}1
đặt left(dfrac{1}{xy};dfrac{1}{yz};dfrac{1}{zx}right)left(a;b;cright)Rightarrow a+b+c1 left(dfrac{ac}{b};dfrac{ab}{c};dfrac{bc}{a}right)left(dfrac{1}{x^2};dfrac{1}{y^2};dfrac{1}{z^2}right)
ta có VTdfrac{1}{sqrt{1+dfrac{1}{x^2}}}+dfrac{1}{sqrt{1+dfrac{1}{y^2}}}+dfrac{1}{sqrt{1+dfrac{1}{z^1}}}sqrt{dfrac{1}{1+dfrac{ac}{b}}}+sqrt{dfrac{1}{1+dfrac{ab}{c}}}+sqrt{dfrac{1}{1+dfrac{bc}{a}}}
dfrac{1}{sqrt{dfrac{b+ac}{b}}}+dfrac{1}{sqrt{dfrac...
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từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
13 tháng 3 2022 lúc 23:29
Cho ba số thực a,b,c dương chứng minh rằng:
\(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}+\sqrt{\dfrac{b^3}{b^3+\left(c+a\right)^3}}+\sqrt{\dfrac{c^3}{c^3+\left(a+b\right)^3}}\ge1\)
cho ba số thực a,b,c dương . chứng minh rằng :
\(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}\)+ \(\sqrt{\dfrac{b^3}{b^3+\left(c+a\right)^3}}\)+\(\sqrt{\dfrac{c^3}{c^3+\left(b+a\right)^3}}\)≥1
bn tham khảo nha
https://hoc24.vn/cau-hoi/cho-ba-so-thuc-abc-duong-chung-minh-rangsqrtdfraca3a3leftbcright3sqrtdfracb3b3leftcaright3sqrtdfracc3c.5222680437292
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Cho a,b,c là các số dương . Chứng minh rằng :
\(\sqrt{\dfrac{a^3}{a^3+\left(b+c\right)^3}}\)+ \(\sqrt{\dfrac{b^3}{b^3+\left(c+a\right)^3}}\) + \(\sqrt{\dfrac{c^3}{c^3+\left(a+b\right)^3}}\)
Cho a, b, c > 0 thoả mãn: \(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\). Chứng minh rằng: \(\dfrac{\sqrt{a}}{a+1}+\dfrac{\sqrt{b}}{b+1}+\dfrac{\sqrt{c}}{c+1}=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
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Cho a,b,c > 0 thỏa abc=1.Chứng minh :
\(P=\dfrac{1}{\sqrt{a\left(1+b\right)}}+\dfrac{1}{\sqrt{b\left(1+c\right)}}+\dfrac{1}{\sqrt{c\left(1+a\right)}}>2\)
\(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)
\(P=\sqrt{\dfrac{yz}{xy+xz}}+\sqrt{\dfrac{zx}{xy+yz}}+\sqrt{\dfrac{xy}{yz+zx}}\)
\(P=\dfrac{2yz}{2\sqrt{yz\left(xy+xz\right)}}+\dfrac{2zx}{2\sqrt{zx\left(xy+yz\right)}}+\dfrac{2xy}{2\sqrt{xy\left(yz+zx\right)}}\)
\(P\ge\dfrac{2yz}{xy+yz+zx}+\dfrac{2zx}{xy+yz+zx}+\dfrac{2xy}{xy+yz+zx}=2\)
Dấu "=" không xảy ra nên \(P>2\)
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Cho a,b,c>0 và a+b+c=căn a +căn b +căn c=2.Tính A=
\(\left(\dfrac{\sqrt{a}}{1+a}+\dfrac{\sqrt{b}}{1+b}+\dfrac{\sqrt{c}}{1+c}\right)\left(\sqrt{1+a}\right)\left(\sqrt{1+b}\right)\left(\sqrt{1+c}\right)\)
Lời giải:
\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)
\(\Rightarrow (\sqrt{a}+\sqrt{b}+\sqrt{c})^2=4\)
\(\Leftrightarrow a+b+c+2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=4\)
\(\Leftrightarrow \sqrt{ab}+\sqrt{bc}+\sqrt{ac}=\frac{4-(a+b+c)}{2}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})\)
Tương tự:
$b+1=(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})$
$c+1=(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})$
Khi đó:
\(A=\left[\frac{\sqrt{a}}{(\sqrt{a}+\sqrt{b})(\sqrt{a}+\sqrt{c})}+\frac{\sqrt{b}}{(\sqrt{b}+\sqrt{a})(\sqrt{b}+\sqrt{c})}+\frac{\sqrt{c}}{(\sqrt{c}+\sqrt{a})(\sqrt{c}+\sqrt{b})}\right]\sqrt{(a+1)(b+1)(c+1)}\)
\(\frac{\sqrt{a}(\sqrt{b}+\sqrt{c})+\sqrt{b}(\sqrt{c}+\sqrt{a})+\sqrt{c}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.\sqrt{(\sqrt{a}+\sqrt{b})^2(\sqrt{b}+\sqrt{c})^2(\sqrt{c}+\sqrt{a})^2}\)
\(=\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})}.(\sqrt{a}+\sqrt{b})(\sqrt{b}+\sqrt{c})(\sqrt{c}+\sqrt{a})\)
\(=2(\sqrt{ab}+\sqrt{bc}+\sqrt{ac})=2\)
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