Tìm min:
\(M=4\left(a+b+c\right)^2+3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) \(với\) \(a,b,c>0\)
Những câu hỏi liên quan
+) Tìm min
\(E=\dfrac{1+\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}{xy+yz+zx}\)
+) Tìm max và min
\(F=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\)
Trong đó a,b,c>0 và \(min\left\{a,b,c\right\}\ge\dfrac{1}{4}max\left\{a,b,c\right\}\)
cho a,b,c>0 và abc=1. Tìm min:
\(Q=\dfrac{a^4}{\left(a^2+b^2\right)\left(a+b\right)}+\dfrac{b^4}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{c^4}{\left(c^2+a^2\right)\left(c+a\right)}\)
Tìm min P= \(\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\left(a^3+b^3+c^3\right)\) biết \(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\) =11 và a,b,c>0.
Cho a,b,c >0 và abc=1. Tìm min:
\(P=\dfrac{a^4+b^4}{\left(a^2+b^2\right)\left(a+b\right)}+\dfrac{b^4+c^4}{\left(b^2+c^2\right)\left(b+c\right)}+\dfrac{a^4+c^4}{\left(a^2+c^2\right)\left(a+c\right)}\)
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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a,b,c>0 . Tìm Min \(E=\left(1+\dfrac{a}{2b}\right)\left(1+\dfrac{b}{2c}\right)\left(1+\dfrac{1}{2a}\right)\)
Số hạng cuối là \(1+\dfrac{c}{2a}\) mới đúng chứ bạn?
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\(E=\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{a}{2b}\right)\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{b}{2c}\right)\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{c}{2a}\right)\)
\(E\ge3\sqrt[3]{\dfrac{a}{8b}}.3\sqrt[3]{\dfrac{b}{8c}}.3\sqrt[3]{\dfrac{c}{8a}}=\dfrac{27}{8}\)
\(E_{min}=\dfrac{27}{8}\) khi \(a=b=c\)
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Giúp vs mọi người ơi
1. a,b,c 0. C/m: dfrac{c^2}{a+b}+dfrac{a^2}{b+c}+dfrac{b^2}{a+c}dfrac{a+b+c}{2}
2. a,b,c 0 và a+b+c 1. C/m: dfrac{1}{a^2+2bc}+dfrac{1}{b^2+2ac}+dfrac{1}{c^2+2ab}9
3. a,b,c là 3 cạnh của một tam giác; pdfrac{a+b+c}{2}
C/m: dfrac{1}{left(p-aright)^2}+dfrac{1}{left(p-bright)^2}+dfrac{1}{left(p-cright)^2}dfrac{p}{left(p-aright)left(p-bright)left(p-cright)}
4. a,b,c 0 và (a+c)(b+c)1
C/m: dfrac{1}{left(a-bright)^2}+dfrac{1}{left(a+cright)^2}+dfrac{1}{left(b+cright)^2}4
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Giúp vs mọi người ơi
1. a,b,c > 0. C/m: \(\dfrac{c^2}{a+b}+\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}>=\dfrac{a+b+c}{2}\)
2. a,b,c > 0 và a+b+c <= 1. C/m: \(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}>=9\)
3. a,b,c là 3 cạnh của một tam giác; \(p=\dfrac{a+b+c}{2}\)
C/m: \(\dfrac{1}{\left(p-a\right)^2}+\dfrac{1}{\left(p-b\right)^2}+\dfrac{1}{\left(p-c\right)^2}>=\dfrac{p}{\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
4. a,b,c > 0 và (a+c)(b+c)=1
C/m: \(\dfrac{1}{\left(a-b\right)^2}+\dfrac{1}{\left(a+c\right)^2}+\dfrac{1}{\left(b+c\right)^2}>=4\)
câu 1: \(VT=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
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Cho \(a,b,c\ge0\) t/m: \(\left\{{}\begin{matrix}c\left(a+b\right)>0\\\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\le6\end{matrix}\right.\)
Tìm Min: \(H=\left(a+b\right)\sqrt{1+\dfrac{3}{a+b^4}}+\sqrt{c^2+\dfrac{3}{c^2}}+\dfrac{\left(b+6\right)^2}{9\left(a+b+c\right)}\)
cho a,b,c > 0 có a+b+c\(\le\)3. Tìm Min
B=\(\dfrac{1}{\left(a+2b\right)\left(a+2c\right)}+\dfrac{1}{\left(b+2a\right)\left(b+2c\right)}+\dfrac{1}{\left(c+2a\right)\left(c+2b\right)}\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(B=\frac{1}{(a+2b)(a+2c)}+\frac{1}{(b+2a)(b+2c)}+\frac{1}{(c+2a)(c+2b)}\)
\(\geq \frac{9}{(a+2b)(a+2c)+(b+2a)(b+2c)+(c+2a)(c+2b)}\)
\(\Leftrightarrow B\geq \frac{9}{(a^2+2ac+2ab+4bc)+(b^2+2bc+2ab+4ac)+(c^2+2bc+2ac+4ab)}\)
\(\Leftrightarrow B\geq \frac{9}{a^2+b^2+c^2+8(ab+bc+ac)}=\frac{9}{(a+b+c)^2+6(ab+bc+ac)}(*)\)
Theo hệ quả quen thuộc của BĐT Cô-si:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow (a+b+c)^2\geq 3(ab+bc+ac)\)
\(\Rightarrow 2(a+b+c)^2\geq 6(ab+bc+ac)(**)\)
Từ \((*); (**)\Rightarrow B\geq \frac{9}{(a+b+c)^2+2(a+b+c)^2}=\frac{3}{(a+b+c)^2}\geq \frac{3}{3^2}=\frac{1}{3}\)
(do \(a+b+c\leq 3)\)
Do đó: \(B_{\min}=\frac{1}{3}\)
Dấu bằng xảy ra khi \(a=b=c=1\)
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cho các số thực dương thỏa mãn \(a+b+c\le\dfrac{3}{2}\)
tìm min \(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{3}{2}\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\dfrac{1}{2}\ge\sqrt[3]{abc}\Rightarrow\dfrac{1}{8}\ge abc\)
Áp dụng BĐT Holder ta có:
\(B=\left(3+\dfrac{1}{a}+\dfrac{1}{b}\right)\left(3+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(3+\dfrac{1}{c}+\dfrac{1}{a}\right)\)
\(\ge\left(\sqrt[3]{3\cdot3\cdot3}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}+\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\right)^3\)
\(=\left(3+2\sqrt[3]{\dfrac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\dfrac{1}{\dfrac{1}{8}}}\right)^3=343\)
Khi \(a=b=c=\dfrac{1}{2}\)
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