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NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 7 2020
4.
\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)
\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)
Dấu "=" xảy ra khi \(a=b=c\)
5.
\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)
Cộng vế với vế:
\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 7 2020
1.
Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)
\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
2.
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Cộng vế với vế:
\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)
3.
Từ câu b, thay \(c=1\) ta được:
\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)
DN
5 tháng 8 2017
Ta có: \(a+b+c\ge3\sqrt[3]{abc}\)
\(ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)(1)
Ta có: \(\left(a-b\right)^3+\left(b-c\right)^2+\left(c-a\right)^3\)
\(=\left(a-b\right)^3+3\left(a-b\right)^2\left(b-c\right)+3\left(a-b\right)\left(b-c\right)^2+\left(b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)^2\left(b-c\right)-3\left(a-b\right)\left(b-c\right)^2\)
\(=\left(a-b+b-c\right)^3-\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Ta có: \(a-b+b-c+c-a\ge3\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge\sqrt[3]{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Leftrightarrow0\ge3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow9abc\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)(2)
Từ (1), (2) ta có: \(\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc+\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
Dấu "=" xảy ra khi \(a=b=c\)
KK
4 tháng 4 2017
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\left\{{}\begin{matrix}3+a^2\ge\left(a+c\right)\left(a+b\right)\\3+b^2\ge\left(a+b\right)\left(b+c\right)\\3+c^2\ge\left(a+c\right)\left(b+c\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{bc}{\sqrt{3+a^2}}\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}\\\dfrac{ca}{\sqrt{3+b^2}}\le\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}\\\dfrac{ab}{\sqrt{3+c^2}}\le\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{bc}{\sqrt{\left(a+c\right)\left(a+b\right)}}+\dfrac{ca}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(\Leftrightarrow VT\le\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}\le\dfrac{\dfrac{bc}{a+c}+\dfrac{bc}{a+b}}{2}\\\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}\le\dfrac{\dfrac{ca}{a+b}+\dfrac{ca}{b+c}}{2}\\\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\dfrac{ab}{a+c}+\dfrac{ab}{b+c}}{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{\left(\dfrac{bc}{a+c}+\dfrac{ab}{a+c}\right)+\left(\dfrac{bc}{a+b}+\dfrac{ca}{a+b}\right)+\left(\dfrac{ab}{b+c}+\dfrac{ca}{b+c}\right)}{2}\)
\(\Rightarrow\sqrt{\dfrac{b^2c^2}{\left(a+c\right)\left(a+b\right)}}+\sqrt{\dfrac{c^2a^2}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{a^2b^2}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\) (2)
Xét \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ca+bc}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (3)
Từ (1) , (2) , (3)
\(\Rightarrow VT\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{bc}{\sqrt{a^2+3}}+\dfrac{ca}{\sqrt{b^2+3}}+\dfrac{ab}{\sqrt{c^2+3}}\le\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=1\)
31 tháng 7 2017
Chứng minh : \(\left(x^2+y^2+z^2\right)^2\ge3\left(x^3y+y^3z+z^3x\right)\)
\(\Leftrightarrow\dfrac{1}{2}\left(\left(x^2-y^2-xy-xz+2yz\right)^2+\left(y^2-z^2-yz-xy+2xz\right)^2+\left(z^2-x^2-xz-yz+2xy\right)^2\right)\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{ab+1}=a-\dfrac{a^2b}{ab+1}\ge a-\dfrac{a^2b}{2\sqrt{ab}}=a-\dfrac{\sqrt{a^3b}}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b}{bc+1}\ge b-\dfrac{\sqrt{b^3c}}{2};\dfrac{c}{ca+1}\ge c-\dfrac{\sqrt{c^3a}}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge3-\dfrac{1}{2}\left(\sqrt{a^3b}+\sqrt{b^3c}+\sqrt{c^3a}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
Xảy ra khi \(a=b=c=1\)
NV
Nguyễn Việt Lâm
Giáo viên
4 tháng 5 2019
\(a^3+a\ge2a^2\) ; \(b^3+b\ge2b^2\); \(c^3+c\ge2c^2\)
\(\Rightarrow a^3+b^3+c^3+3\ge2\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^3+b^3+c^3+ab+ac+bc\ge2\left(a^2+b^2+c^2\right)+ab+ac+bc-3\)
Mặt khác
\(P=2\left(a^2+b^2+c^2\right)+ab+ac+bc-3=\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{1}{2}\left(a+b+c\right)^2-3\)
\(P=\frac{3}{2}\left(a^2+b^2+c^2\right)+\frac{3}{2}\ge\frac{3}{2}.\frac{\left(a+b+c\right)^2}{3}+\frac{3}{2}=6\)
\(\Rightarrow a^3+b^3+c^3+ab+ac+bc\ge6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
KK
24 tháng 5 2018
Ta có \(\dfrac{a^2}{a+b^2}=a-\dfrac{ab^2}{a+b^2}\ge a-\dfrac{ab^2}{2b\sqrt{a}}=a-\dfrac{ab}{2\sqrt{a}}\)
Thiết lập tương tự và thu lại ta có :
\(VT\ge3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\)
Xét \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}=\sqrt{\dfrac{a^2b^2}{4a}}+\sqrt{\dfrac{b^2c^2}{4b}}+\sqrt{\dfrac{a^2c^2}{4c}}\)
Áp dụng bđt Cauchy ta có \(\sqrt{\dfrac{a^2b^2}{4a}}=\sqrt{\dfrac{ab}{2a}.\dfrac{ab}{2}}\le\dfrac{\dfrac{b}{2}+\dfrac{ab}{2}}{2}\)
Thiết lập tương tự và thu lại ta có :
\(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{\dfrac{a+b+c}{2}+\dfrac{ab+bc+ac}{2}}{2}=\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\left(1\right)\)
Theo hệ quả của bđt Cauchy ta có \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
\(\Rightarrow ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow\dfrac{\dfrac{3}{2}+\dfrac{ab+bc+ac}{2}}{2}\le\dfrac{\dfrac{3}{2}+\dfrac{3}{2}}{2}=\dfrac{3}{2}\left(2\right)\)
Từ ( 1 ) và ( 2 ) ta có \(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\le\dfrac{3}{2}\)
\(\Rightarrow3-\left(\dfrac{ab}{2\sqrt{a}}+\dfrac{bc}{2\sqrt{b}}+\dfrac{ac}{2\sqrt{c}}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\)
\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)
Dấu '' = '' xảy ra khi \(a=b=c=1\)
HN
24 tháng 4 2017
Ta có: \(\left\{{}\begin{matrix}3\sqrt{3}a^2+\sqrt{a}+\sqrt{a}\ge3\sqrt{3}a\left(1\right)\\3\sqrt{3}b^2+\sqrt{b}+\sqrt{b}\ge3\sqrt{3}b\left(2\right)\\3\sqrt{3}c^2+\sqrt{c}+\sqrt{c}\ge3\sqrt{3}c\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được
\(2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge3\sqrt{3}\left[\left(a+b+c\right)-\left(a^2+b^2+c^2\right)\right]\)
\(\Leftrightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}\ge\dfrac{3\sqrt{3}\left[1-\left(a^2+b^2+c^2\right)\right]}{2}\)
\(=\dfrac{3\sqrt{3}\left[1-\left(a+b+c\right)^2+2\left(ab+bc+ca\right)\right]}{2}\)
\(=3\sqrt{3}\left(ab+bc+ca\right)\)
\(\RightarrowĐPCM\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 2 2020
1.
\(6=\frac{\sqrt{2}^2}{x}+\frac{\sqrt{3}^2}{y}\ge\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{x+y}=\frac{5+2\sqrt{6}}{x+y}\)
\(\Rightarrow x+y\ge\frac{5+2\sqrt{6}}{6}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\frac{x}{\sqrt{2}}=\frac{y}{\sqrt{3}}\\x+y=\frac{5+2\sqrt{6}}{6}\end{matrix}\right.\)
Bạn tự giải hệ tìm điểm rơi nếu thích, số xấu quá
2.
\(VT\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
Đặt \(x+y+z=t\Rightarrow0< t\le1\)
\(VT\ge\sqrt{t^2+\frac{81}{t^2}}=\sqrt{t^2+\frac{1}{t^2}+\frac{80}{t^2}}\ge\sqrt{2\sqrt{\frac{t^2}{t^2}}+\frac{80}{1^2}}=\sqrt{82}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 2 2020
3.
\(\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{a^2}{b^5}+\frac{1}{a^3}+\frac{1}{a^3}\ge5\sqrt[5]{\frac{a^6}{b^{15}.a^6}}=\frac{5}{b^3}\)
Tương tự: \(\frac{3b^2}{c^5}+\frac{2}{b^3}\ge\frac{5}{a^3}\) ; \(\frac{3c^2}{d^5}+\frac{2}{c^3}\ge\frac{5}{d^3}\) ; \(\frac{3d^2}{a^5}+\frac{2}{d^2}\ge\frac{5}{a^3}\)
Cộng vế với vế và rút gọn ta được: \(3VT\ge3VP\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=1\)
4.
ĐKXĐ: \(-2\le x\le2\)
\(y^2=\left(x+\sqrt{4-x^2}\right)^2\le2\left(x^2+4-x^2\right)=8\)
\(\Rightarrow y\le2\sqrt{2}\Rightarrow y_{max}=2\sqrt{2}\) khi \(x=\sqrt{2}\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge-2\\\sqrt{4-x^2}\ge0\end{matrix}\right.\) \(\Rightarrow x+\sqrt{4-x^2}\ge-2\)
\(y_{min}=-2\) khi \(x=-2\)
PD
1
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 2 2020
\(P=\frac{b^2c^2+c^2a^2+a^2b^2}{abc}\Rightarrow P^2=\frac{b^4c^4+c^4a^4+a^4b^4+2a^2b^2c^2\left(a^2+b^2+c^2\right)}{a^2b^2c^2}\)
\(P^2\ge\frac{a^2b^2c^2\left(a^2+b^2+c^2\right)+2a^2b^2c^2}{a^2b^2c^2}=\frac{3a^2b^2c^2}{a^2b^2c^2}=3\)
\(\Rightarrow P\ge\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Ta có:
\(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
\(\Rightarrow VT-VP=a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\) (Giải thích:\(-6\ge-2\left(ab+bc+ca\right)\Rightarrow a^3+b^3+c^3+ab+bc+ca-6\ge a^3+b^3+c^3-ab-bc-ca\))
Ta lại có:
\(a^3+b^3+c^3-ab-bc-ca\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}-\frac{\left(a+b+c\right)^2}{3}\ge\frac{\left[\frac{\left(a+b+c\right)^2}{3}\right]^2}{3}-3=0\)
\(\Rightarrow VT-VP\ge0\)
\(\Rightarrow P\ge6\)
Nếu có không đúng thì nhớ nói nhe chớ đừng có k sai tui giống mấy lần trước nhe :(
Bài ở dưới mình nhầm nhe.
Update
Ta có:
\(a^3+b^3+c^3\ge\frac{\left(a^2+b^2+c^2\right)^2}{a+b+c}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{3}\ge\frac{\left(a^2+b^2+c^2\right)\frac{\left(a+b+c\right)^2}{3}}{3}=a^2+b^2+c^2\)
\(\Rightarrow P\ge a^2+b^2+c^2+ab+bc+ca=\frac{a^2+b^2+c^2}{2}+\frac{\left(a+b+c\right)^2}{2}\ge\frac{\frac{\left(a+b+c\right)^2}{3}}{2}+\frac{9}{2}=6\)