Cho 2 số dương a,b thỏa mãn a+b=4. tìm GTNN của A=\(\frac{a^2+b^2}{4}\)
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N
11 tháng 9 2019
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
N
11 tháng 9 2019
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
MN
0
6 tháng 12 2017
\(Ta có: \frac{{a^5 }}{{b^3 + c^2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }}\mathop \ge \frac{{3a^2 }}{2}\)
\(\Rightarrow \frac{{a^5 }}{{b^3 + c^2 }} \ge \frac{{3a^2 }}{2} - (\frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }} + \frac{{\sqrt {a(b^3 + c^2 )} }}{{2\sqrt 2 }})\)
\(Do đó: \frac{{a^5 }}{{b^3 + c^2 }} \ge \frac{{3a^2 }}{2} - \frac{{\sqrt {2a(b^3 + c^2 )} }}{2}\mathop \ge \frac{{3a^2 }}{2} - \frac{{2a + b^3 + c^2 }}{4}\)
\(CMTT \frac{{b^5 }}{{c^3 + a^2 }}\mathop \ge \frac{{3b^2 }}{2} - \frac{{2b + c^3 + a^2 }}{4}\), \(\frac{{c^5}}{{a^3+b^2}}\mathop \ge \frac{{3c^2 }}{2} - \frac{{2c + a^3 + b^2 }}{4}\)
\(M \ge \frac{{3(a^2 + b^2 + c^2 )}}{2} + a^4 + b^4 + c^4 - \frac{{2(a + b + c) + (a^2 + b^2 + c^2 ) + (a^3 + b^3 + c^3 )}}{4}\)
\(M \ge \frac{9}{2} + a^4 + b^4 + c^4 - \frac{{2(a + b + c) + (a^2 + b^2 + c^2 ) + (a^3 + b^3 + c^3 )}}{4}\)
Áp dụng Bunhiacoopski ta có:
\(\sqrt {(a^4+b^4+c^4 )(a^2+b^2+c^2)}=\sqrt {(a^4 +b^4+ c^4 ).3}\ge a^3+b^3+c^3 \)
\(\sqrt {(a^4 + b^4 + c^4 )(1 + 1 + 1)} = \sqrt {(a^2 + b^2 + c^2 ).3} \ge a^2 + b^2 + c^2 \Leftrightarrow a^4 + b^4 + c^4 \ge 3\)
Ta có: \(3 = a^2 + b^2 + c^2 \ge \frac{{(a + b + c)^2 }}{3} \Leftrightarrow a^2 + b^2 + c^2 \ge a + b + c\)
\(Đặt t=x^4+y^4+z^4 (t \ge 3) cần CM để trở thành S \ge \frac{{4t - 9 - \sqrt {3t} }}{4}\ge 0\)
\(Ta có: S\ge \frac{{4t - 9 - \sqrt {3t} }}{4} = \frac{{3(t - 3) + \sqrt t (\sqrt t - \sqrt 3 )}}{4} \ge 0
\)
\(Do đó: M\geq \frac{9}{2}\)
6 tháng 12 2017
Phần đầu mình thiếu nha
\(\frac{a^5}{b^3+c^2}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}\ge\frac{3a^2}{2}\)
=> \(\frac{a^5}{b^3+c^2}\ge\frac{3a^2}{2}-\left(\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}+\frac{\sqrt{a\left(b^3+c^2\right)}}{2\sqrt{2}}\right)\)
Do đó \(\frac{a^5}{b^3+c^2}\ge\frac{3a^2}{2}-\frac{\sqrt{2a\left(b^3+c^2\right)}}{2}\ge\frac{3a^2}{2}-\frac{\left(2a+b^3+b^2\right)}{4}\)
CMTT \(\frac{b^5}{c^3+a^2}\ge\frac{3b^2}{2}-\frac{\left(2b+c^3+a^2\right)}{4},\frac{c^5}{a^3+b^2}\ge\frac{3c^2}{2}-\frac{\left(2c+a^3+b^2\right)}{4}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 3 2022
Áp dụng BĐT Cô-si:
\(a^2+3\ge2\sqrt{3a^2}=2\sqrt{3}a\)
Tương tự: \(b^2+3\ge2\sqrt{3}b\) ; \(c^2+3\ge2\sqrt{3}c\)
Cộng vế: \(a^2+b^2+c^2+9\ge2\sqrt{3}\left(a+b+c\right)\)
\(\Rightarrow a+b+c\le\dfrac{a^2+b^2+c^2+9}{2\sqrt{3}}=\dfrac{9+9}{2\sqrt{3}}=3\sqrt{3}\)
\(\Rightarrow-\left(a+b+c\right)\ge-3\sqrt{3}\)
Tiếp tục áp dụng BĐT Cô-si:
\(\dfrac{a^4}{b+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(b+2\right)\ge2\sqrt{\dfrac{9a^4\left(b+2\right)}{\left(b+2\right)\left(2+\sqrt{3}\right)^2}}=\dfrac{6a^2}{2+\sqrt{3}}\)
Tương tự:
\(\dfrac{b^4}{c+2}+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(c+2\right)\ge\dfrac{6b^2}{2+\sqrt{3}}\)
\(\dfrac{c^4}{a+2}+\dfrac{9}{\left(2+\sqrt{3}\right)}\left(a+2\right)\ge\dfrac{6c^2}{2+\sqrt{3}}\)
Cộng vế:
\(P+\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{6}{2+\sqrt{3}}\left(a^2+b^2+c^2\right)=\dfrac{54}{2+\sqrt{3}}\)
\(\Rightarrow P\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}\left(a+b+c+6\right)\ge\dfrac{54}{2+\sqrt{3}}-\dfrac{9}{\left(2+\sqrt{3}\right)^2}.\left(3\sqrt{3}+6\right)\)
\(\Rightarrow P\ge\dfrac{27}{2+\sqrt{3}}=27\left(2-\sqrt{3}\right)\)
Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)
\(4^2=\left(a+b\right)^2\le\left(1^2+1^2\right)\left(a^2+b^2\right)\)\(\Leftrightarrow\left(a^2+b^2\right)\ge8\)\(\Leftrightarrow A=\frac{a^2+b^2}{4}\ge\frac{8}{4}=2\)
Min A = 2 khi a =b =2