Chứng minh rằng:
a, \(4\left(a^3+b^3\right)\ge\left(a+b\right)^3\) với a> b> 0
b, \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
c, \(\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^3\)
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NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2022
\(\dfrac{a^3}{\left(b+2\right)\left(c+3\right)}+\dfrac{b+2}{36}+\dfrac{c+3}{48}\ge3\sqrt[3]{\dfrac{a^3\left(b+2\right)\left(c+3\right)}{1728\left(b+2\right)\left(c+3\right)}}=\dfrac{a}{4}\)
Tương tự: \(\dfrac{b^3}{\left(c+2\right)\left(a+3\right)}+\dfrac{c+2}{36}+\dfrac{a+3}{48}\ge\dfrac{b}{4}\)
\(\dfrac{c^3}{\left(a+2\right)\left(b+3\right)}+\dfrac{a+2}{36}+\dfrac{b+3}{48}\ge\dfrac{c}{4}\)
Cộng vế:
\(P+\dfrac{7\left(a+b+c\right)}{144}+\dfrac{17}{48}\ge\dfrac{a+b+c}{4}\)
\(\Rightarrow P\ge\dfrac{29}{144}\left(a+b+c\right)-\dfrac{17}{48}\ge\dfrac{29}{144}.3\sqrt[3]{abc}-\dfrac{17}{48}=\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
LD
29 tháng 12 2018
a.
\(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)
\(\Leftrightarrow a^4+b^4\ge ab^3+a^3b\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(*)
Mà \(a^2+ab+b^2=\left(a^2+2\cdot a\cdot\dfrac{1}{2}b+\dfrac{b^2}{4}\right)+\dfrac{3b^2}{4}=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\)
Suy ra (*) đúng => đpcm
Dấu "=" xảy ra khi a = b
LD
29 tháng 12 2018
b.
\(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(\Leftrightarrow3a^4+3b^4+3c^4\ge a^4+ab^3+ac^3+a^3b+b^4+bc^3+a^3c+b^3c+c^4\)
\(\Leftrightarrow2a^4+2b^4+2c^4\ge ab^3+a^3b+b^3c+bc^3+ca^3+c^3a\)
\(\Leftrightarrow\left(a^4+b^4\right)+\left(b^4+c^4\right)+\left(c^4+a^4\right)\ge\left(a^3b+ab^3\right)+\left(b^3c+bc^3\right)+\left(c^3a+ca^3\right)\)
Theo câu a. thì điều này đúng
Dấu "=" khi a=b=c
GH
18 tháng 6 2023
Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)
Ta có:
\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)
BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)
Đánh giá cuối cùng đúng theo BĐT Cauchy
Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi a = b = c = 1.
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2021
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
AH
Akai Haruma
Giáo viên
19 tháng 3 2018
Lời giải:
Sử dụng pp biến đổi tương đương:
a) \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\)
\(\Leftrightarrow \frac{a^2+b^2}{2}\geq \frac{(a+b)^2}{4}\)
\(\Leftrightarrow 4(a^2+b^2)\geq 2(a+b)^2\Leftrightarrow 4(a^2+b^2)\geq 2(a^2+2ab+b^2)\)
\(\Leftrightarrow 2(a^2+b^2)\geq 4ab\Leftrightarrow 2(a^2+b^2-2ab)\geq 0\)
\(\Leftrightarrow 2(a-b)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xẩy ra khi $a=b$
c)
\(\frac{a^2+b^2+c^2}{3}\geq \left(\frac{a+b+c}{3}\right)^2\) \(\Leftrightarrow \frac{a^2+b^2+c^2}{3}\geq \frac{(a+b+c)^2}{9}\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq (a+b+c)^2\)
\(\Leftrightarrow 3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)\)
\(\Leftrightarrow 2(a^2+b^2+c^2)\geq 2(ab+bc+ac)\)
\(\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ac+a^2)\geq 0\)
\(\Leftrightarrow (a-b)^2+(b-c)^2+(c-a)^2\geq 0\) (luôn đúng)
Do đó ta có đpcm. Dấu bằng xảy ra khi $a=b=c$
b) \(\frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\)
Áp dụng 2 lần BĐT phần a: \(\frac{a^4+b^4}{2}\geq \left(\frac{a^2+b^2}{2}\right)^2(1)\)
Và: \(\frac{a^2+b^2}{2}\geq \left(\frac{a+b}{2}\right)^2\Rightarrow \left(\frac{a^2+b^2}{2}\right)^2\geq \left(\frac{a+b}{2}\right)^4(2)\)
Từ \((1); (2)\Rightarrow \frac{a^4+b^4}{2}\geq \left(\frac{a+b}{2}\right)^4\) (đpcm)
Dấu bằng xảy ra khi \(a=b\)
AH
Akai Haruma
Giáo viên
18 tháng 9 2021
Lời giải:
Áp dụng BĐT AM-GM:
\(\frac{a^5}{b^2(c+3)}+\frac{b(c+3)}{16}+\frac{ab}{4}\geq \frac{3}{4}a^2\)
Tương tự với các phân thức còn lại và cộng theo vế:
\(A+\frac{5}{16}ab+\frac{3(a+b+c)}{16}\geq \frac{3}{4}(a^2+b^2+c^2)\)
Mà theo BĐT AM-GM dễ thấy \(a^2+b^2+c^2\geq ab+bc+ac\Rightarrow A\geq \frac{7}{16}(a^2+b^2+c^2)-\frac{3}{16}(a+b+c)\)
Áp dụng BĐT AM-GM tiếp:
$a^2+1\geq 2a; b^2+1\geq 2b; c^2+1\geq 2c$
$\Rightarrow a^2+b^2+c^2+3\geq 2(a+b+c)\geq a+b+c+3\sqrt[3]{abc}=a+b+c+3$
$\Rightarrow a^2+b^2+c^2\geq a+b+c\Rightarrow A\geq \frac{1}{4}(a+b+c)\geq \frac{1}{4}\sqrt[3]{abc}=\frac{3}{4}$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
AH
Akai Haruma
Giáo viên
18 tháng 9 2021
Mình vừa sửa lỗi công thức, bạn load lại để xem nhé.
KK
7 tháng 5 2017
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\dfrac{a^2}{2}+\dfrac{b^2}{c}+\dfrac{c^2}{c}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)
\(\Leftrightarrow a^2-\dfrac{a^2}{2}+b^2-\dfrac{b^2}{2}+c^2-\dfrac{c^2}{2}\ge\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ca}{2}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{2\left(a^2+b^2+c^2+ab+bc+ca\right)}{4}\)
\(\Leftrightarrow a^2+b^2+c^2\ge\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{4}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(\Rightarrow\left(a+b\right)^2\ge4ab\)
Tương tự ta có \(\left\{{}\begin{matrix}\left(b+c\right)^2\ge4bc\\\left(c+a\right)^2\ge4ca\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^2c+\left(a+b\right)^2\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2a+\left(b+c\right)^2\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2b+\left(c+a\right)^2\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2\left(c+1\right)\ge4abc+\left(a+b\right)^2\\\left(b+c\right)^2\left(a+1\right)\ge4abc+\left(b+c\right)^2\\\left(c+a\right)^2\left(b+1\right)\ge4abc+\left(c+a\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}\le\dfrac{8}{4abc+\left(a+b\right)^2}\\\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}\le\dfrac{8}{4abc+\left(b+c\right)^2}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}\le\dfrac{8}{4abc+\left(c+a\right)^2}\end{matrix}\right.\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\) (3)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{\left(a+b\right)^2}{4}\ge2\sqrt{\dfrac{2}{c+1}}=\dfrac{4}{\sqrt{2\left(c+1\right)}}\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{\left(b+c\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(a+1\right)}}\\\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(c+a\right)^2}{4}\ge\dfrac{4}{\sqrt{2\left(b+1\right)}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2\left(c+1\right)}+\dfrac{8}{\left(b+c\right)^2\left(a+1\right)}+\dfrac{8}{\left(c+a\right)^2\left(b+1\right)}+\dfrac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a^2\right)}{4}\ge\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\)(4)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\sqrt{2\left(c+1\right)}\le\dfrac{c+3}{2}\)
\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}\ge\dfrac{8}{c+3}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2\left(a+1\right)}}\ge\dfrac{8}{a+3}\\\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{b+3}\end{matrix}\right.\)
\(\Rightarrow\dfrac{4}{\sqrt{2\left(c+1\right)}}+\dfrac{4}{\sqrt{2\left(a+1\right)}}+\dfrac{4}{\sqrt{2\left(b+1\right)}}\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) (5)
Từ điều (3) , (4) , (5)
\(\Rightarrow\dfrac{8}{\left(a+b\right)^2+4abc}+\dfrac{8}{\left(b+c\right)^2+4abc}+\dfrac{8}{\left(c+a\right)^2+4abc}+a^2+b^2+c^2\ge\dfrac{8}{a+3}+\dfrac{8}{b+3}+\dfrac{8}{c+3}\) ( đpcm )
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 9 2021
\(\dfrac{a^5}{b^2\left(c+3\right)}+\dfrac{b^2}{4}+\dfrac{a\left(c+3\right)}{16}\ge3\sqrt[3]{\dfrac{a^6b^2\left(c+3\right)}{64b^2\left(c+3\right)}}=\dfrac{3}{4}a^2\)
Tương tự: \(\dfrac{b^5}{c^2\left(a+3\right)}+\dfrac{c^2}{4}+\dfrac{b\left(a+3\right)}{16}\ge\dfrac{3}{4}b^2\)
\(\dfrac{c^5}{a^2\left(b+3\right)}+\dfrac{a^2}{4}+\dfrac{c\left(b+3\right)}{16}\ge\dfrac{3}{4}c^2\)
Cộng vế:
\(A+\dfrac{a^2+b^2+c^4}{4}+\dfrac{ab+bc+ca}{16}+\dfrac{9}{16}\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)
\(\Rightarrow A\ge\dfrac{1}{2}\left(a^2+b^2+c^2\right)-\dfrac{ab+bc+ca}{16}-\dfrac{9}{16}\ge\dfrac{1}{2}\left(a^2+b^2+c^2\right)-\dfrac{a^2+b^2+c^2}{16}-\dfrac{9}{16}\)
\(\Rightarrow A\ge\dfrac{7}{16}\left(a^2+b^2+c^2\right)-\dfrac{9}{16}\ge\dfrac{7}{16}.3\sqrt[3]{\left(abc\right)^2}-\dfrac{9}{16}=\dfrac{3}{4}\) (đpcm)
a) Theo bđt cauchy ta có:
\(a^3+b^3+b^3\ge3\sqrt[3]{a^3.b^6}=3ab^2\)
\(a^3+a^3+b^3\ge3a^2b\)
công vế theo vế ta có \(3\left(a^3+b^3\right)\ge3ab^2+3a^2b\)
\(\Leftrightarrow a^3+b^3+3\left(a^3+b^3\right)\ge a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)
suy ra đpcm
ta luôn có \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2+a^2+b^2\ge a^2+2ab+b^2\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow\dfrac{2\left(a^2+b^2\right)}{4}\ge\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow\dfrac{\left(a^2+b^2\right)}{2}\ge\dfrac{\left(a+b\right)^2}{2^2}=\left(\dfrac{a+b}{2}\right)^2\)
suy ra đpcm