Cho \(\hept{\begin{cases}a,b>0\\a+b\le1\end{cases}}\)Tìm Min \(A=\frac{1}{ab}+\frac{1}{a^2+b^2}\)
Cho \(\hept{\begin{cases}ab+1\le b\\a,b>0\end{cases}}\)
Tìm MIN A= \(\frac{a^2+b^2}{ab}\)
1, Cho \(\hept{\begin{cases}a,b>0\\a^2+b^2=1\end{cases}.}\)Tìm min A= \(\left(1+a\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)\)
2, Cho \(\hept{\begin{cases}a^2+2b^2\le3c^2\\a,b,c>0\end{cases}}\).Chứng minh : \(\frac{1}{a}+\frac{2}{b}\ge\frac{3}{c}\)
1,
\(A=1+a+\frac{1}{b}+\frac{a}{b}+1+b+\frac{1}{a}+\frac{b}{a}\)
\(\ge1+1+2\sqrt{\frac{a}{b}.\frac{b}{a}}+a+b+\frac{a+b}{ab}=4+a+b+\frac{4\left(a+b\right)}{\left(a+b\right)^2}=4+a+b+\frac{4}{a+b}\)
lại có \(\left(1+1\right)\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a+b\le\sqrt{2}\)
\(4+a+b+\frac{4}{a+b}=4+\left(a+b+\frac{2}{a+b}\right)+\frac{2}{a+b}\ge4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)
\(\Rightarrow A\ge4+3\sqrt{2}\)
câu 2
ta có:\(\left(2b^2+a^2\right)\left(2+1\right)\ge\left(2b+a\right)^2\Rightarrow3c\ge a+2b\)
\(\frac{1}{a}+\frac{2}{b}=\frac{1}{a}+\frac{4}{2b}\ge\frac{9}{a+2b}\ge\frac{9}{3c}=\frac{3}{c}\left(Q.E.D\right)\)
Cho \(\hept{\begin{cases}a+b=1\\a,b>0\end{cases}}\)
Tìm MIN A=\(a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\)
Vì a,b >0
Áp dụng bất đẳng thức Cauchy, ta có:
\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2.\frac{1}{a^2}}\)
\(\ge2\)
\(b^2+\frac{1}{b^2}\ge2\sqrt{b^2.\frac{1}{b^2}}\)
\(\ge2\)
Cộng vế theo vế, ta được:
\(a^2+b^2+\frac{1}{a^2}+\frac{1}{b^2}\ge2+2\)
\(\Rightarrow A\ge4\)
Vậy MinA=4 \(\Leftrightarrow\orbr{\begin{cases}a^2=\frac{1}{a^2}\\b^2=\frac{1}{b^2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a=1\\b=1\end{cases}}\)
Cho\(\hept{\begin{cases}a,b,c>0\\a+b+c=1\end{cases}}\) Tìm Min\(A=\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}\)
\(A=\frac{1}{a^2+b^2+c^2}+\frac{1}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{a+b+c}{abc}=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\)
\(>=\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+ac+bc}\)(bđt svacxo)\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+ac+bc}+\frac{1}{ab+ac+bc}+\frac{7}{ab+ac+bc}\)
\(>=\frac{9}{a^2+b^2+c^2+ab+ac+bc+ac+ac+bc}+\frac{7}{ab+ac+bc}\)(bđt svacxo)
\(=\frac{9}{a^2+b^2+c^2+2ab+2ac+2bc}+\frac{7}{ab+ac+bc}=\frac{9}{\left(a+b+c\right)^2}+\frac{7}{ab+ac+bc}\)
\(=\frac{9}{1}+\frac{7}{ab+ac+bc}=9+\frac{7}{ab+ac+bc}\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc>=ab+ac+bc+2ab+2ac+2bc\)
\(=3ab+3ac+3bc=3\left(ab+ac+bc\right)\Rightarrow\frac{1}{3}\left(a+b+c\right)^2=\frac{1}{3}\cdot1=\frac{1}{3}>=ab+ac+bc\Rightarrow ab+ac+bc< =\frac{1}{3}\)
\(\Rightarrow9+\frac{7}{ab+ac+bc}>=9+\frac{7}{\frac{1}{3}}=9+7\cdot3=9+21=30\)
\(\Rightarrow A>=30\)dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
vậy min A là 30 khi \(a=b=c=\frac{1}{3}\)
Cho \(\hept{\begin{cases}x+y\le1\\x,y>0\end{cases}}\)
Tìm MIN A=\(\left(\frac{1}{x}+\frac{1}{y}\right)\sqrt{1+x^2y^2}\)
Vì a,b>0
A\(\ge2\sqrt{\frac{1}{x}\cdot\frac{1}{y}}\cdot\sqrt{1+x^2y^2}\)
A\(\ge2\sqrt{\frac{1+x^2y^2}{xy}}\)
A\(\ge2\sqrt{\frac{1}{xy}+xy}\)
Đặt xy=a, a>0
Ta cs xy\(\le\frac{\left(x+y\right)^2}{4}\le\frac{1^2}{4}=\frac{1}{4}\)
ĐK 0<a<\(\frac{1}{4}\)
\(\Leftrightarrow A\ge2\sqrt{\frac{1}{a}+a}\)
A\(\ge2\sqrt{16a+\frac{1}{a}-15a}\)
a>0, áp dụng bđt cô si
\(A\ge2\sqrt{2\sqrt{16a\cdot\frac{1}{a}}-\frac{15}{4}}\)
A\(\ge\sqrt{17}\)
Dấu = x ra a=b=0.5
Cho \(\hept{\begin{cases}a,b,c>0\\a^2+b^2+c^2=1\end{cases}}\)Tìm Min \(B=\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\)
Em không chắc đâu nha, sai thì xin thông cảm cho ạ
\(a=b=c=\frac{\sqrt{3}}{3}\Rightarrow B=\frac{3\sqrt{3}}{2}\). Ta se chung minh do la gia tri min cua B. That vay:
\(BĐT\Leftrightarrow\frac{a}{b^2+c^2}+\frac{b}{c^2+a^2}+\frac{c}{a^2+b^2}\ge\frac{3\sqrt{3}}{2}=\frac{3\sqrt{3}}{2\sqrt{a^2+b^2+c^2}}\)
BĐT trên đồng bậc, nên ta chuẩn hóa a2 + b2 + c2 = 3 và chứng minh:
\(\frac{a}{3-a^2}+\frac{b}{3-b^2}+\frac{c}{3-c^2}\ge\frac{3}{2}\) (2)
Ta chứng minh BĐT sau: \(\frac{a}{3-a^2}\ge\frac{1}{2}a^2\Leftrightarrow\frac{a^2}{2}-\frac{a}{3-a^2}\le0\)
\(\Leftrightarrow\frac{-\left(a-1\right)^2a\left(a+2\right)}{2\left(3-a^2\right)}\le0\) (Đúng)
Tương tự với hai BĐT còn lại và cộng theo vế suy ra BĐT (2) là đúng.
Suy ra BĐT (1) là đúng suy ra \(B_{min}=\frac{3\sqrt{3}}{2}\)
Vậy...
Xét \(\frac{a}{b^2+c^2}=\frac{a}{1-a^2}\ge\frac{3\sqrt{3}}{2}a^2\)
<=> \(a^4-a^2+\frac{2\sqrt{3}}{9}a\ge0\)
<=> \(a\left(a+\frac{2\sqrt{3}}{3}\right)\left(a-\frac{\sqrt{3}}{3}\right)^2\ge0\)luôn đúng
=> \(B\ge\frac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\frac{3\sqrt{3}}{2}\)
Min \(B=\frac{3\sqrt{3}}{2}\)khi \(a=b=c=\frac{\sqrt{3}}{3}\)
Cho \(\hept{\begin{cases}ab+bc+ca\le abc\\a,b,c>0\end{cases}}\)
Tìm Min \(A=\frac{a^2}{b+2a}+\frac{b^2}{c+2b}+\frac{c^2}{a+2c}\)
Theo gt \(ab+bc+ca\le abc^{\left(3\right)}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)
\(\frac{9}{a+b+c}\le1\)
\(a+b+c\ge9^{\left(1\right)}\)
Mặt khác
\(a^2+b^2+c^2\ge3\left(a+b+c\right)\)
\(a^2+b^2+c^2\ge9\cdot3=27^{\left(2\right)}\)
Vì a,b,c >0, áp dụng bất đẳng thức cô si ta có:
\(\frac{a^2b}{b+2a}+\frac{b\left(b+2a\right)}{9}\ge2\sqrt{\frac{a^2b}{b+2a}\cdot\frac{b\left(b+2a\right)}{9}}=\frac{2ab}{3}\)
CMTT
\(\frac{b^2c}{c+2b}+\frac{c\left(c+2b\right)}{9}\ge\frac{2bc}{3}\)
\(\frac{c^2a}{a+2c}+\frac{a\left(a+2c\right)}{9}\ge\frac{2ca}{3}\)
Cộng vế với vế a được :
\(A+\frac{a^2+b^2+c^2}{9}+\frac{2\left(ab+bc+ca\right)}{9}\ge\frac{2\left(ab+bc+ca\right)}{3}\)
\(A\ge\frac{4\left(ab+bc+ca\right)}{3}-\frac{a^2+b^2+c^2}{9}^{\left(#\right)}\)
Từ 1,2,3 và # ta có
\(A\ge\frac{4\cdot9}{3}-\frac{27}{9}=9\)
Dấu bằng xảy ra \(\Leftrightarrow a=b=c=3\)
Vậy...
1. cho \(-1\le a,b,c\le2\) và a+b+c=0. CMR \(a^2+b^2+c^2\le6\)
2. cho \(\hept{\begin{cases}a,b,c>0\\a+b+c=1\end{cases}}\)cmr hoán vị của \(a\sqrt[3]{1+b-c}\ge\frac{3\sqrt{17}}{2}\)
3. \(\hept{\begin{cases}a,b,c>0\\a+b+c=1\end{cases}}\)cmr: hoán vị của\(\frac{a}{a^2+1}\le\frac{9}{10}\)
4. \(\hept{\begin{cases}a,b,c>0\\a+b+c\le\frac{3}{2}\end{cases}}\)cmr: hoán vị của \(a\sqrt[3]{1+b-c}\le1\)
1.
\(-1\le a\le2\Rightarrow\hept{\begin{cases}a+1\ge0\\a-2\le0\end{cases}\Rightarrow\left(a+1\right)\left(a-2\right)\le0\Leftrightarrow a^2\le}2+a\)
Tương tự \(b^2\le2+b,c^2\le2+c\Rightarrow a^2+b^2+c^2\le6+a+b+c=6\)
Dấu "=" xảy ra khi a=2,b=c=-1 và các hoán vị của chúng
Xét \(\frac{a^2+1}{a}=a+\frac{1}{a}\)
Dễ thấy dấu "=" xảy ra khi \(a=\frac{1}{3}\)
khi đó \(a+\frac{1}{a}=a+\frac{1}{9a}+\frac{8}{9a}\ge2\sqrt{\frac{a.1}{9a}}+\frac{8}{\frac{9.1}{3}}=\frac{10}{3}\)
\(\Rightarrow\frac{a}{a^2+1}\le\frac{3}{10}\)
tương tự =>đpcm
lười quá khỏi nghĩ đưa link
| Inequalities (ko dịch dc thì pm)
Cho \(\hept{\begin{cases}a,b,c>0\\a+b+c=3\end{cases}}\)Tìm min \(A=3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(A\ge3\left(a+b+c\right)+\frac{9}{a+b+c}=3.3+\frac{9}{3}=12\)
\(A_{min}=12\) khi \(a=b=c=1\)
Ta cần chứng minh: \(3a+\frac{1}{a}\ge2a+2\Leftrightarrow3a+\frac{1}{a}-4\ge2\left(a-1\right)\)
\(\Leftrightarrow\frac{3a^2-4a+1}{a}-2\left(a-1\right)\ge0\Leftrightarrow\left(a-1\right)\left(\frac{3a-1}{a}-2\right)\ge0\Leftrightarrow\frac{\left(a-1\right)^2}{a}\)(đúng)
Tương tự: \(3b+\frac{1}{b}\ge2b+2;3c+\frac{1}{c}\ge2c+2\)
Cộng theo vế: \(A\ge2\left(a+b+c\right)+6=12\)
Dấu bằng xảy ra khi a=b=c=1
\(A=3\left(a+b+c\right)+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3^2+\frac{9}{a+b+c}=12.\)(BĐT Bunhia)
Dấu "=" xra khi a=b=c=1