MT
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Khách
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MT
0
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8 2021
\(\dfrac{a^5}{b^3+c^2}+\dfrac{b^3+c^2}{4}+\dfrac{a^4}{2}\ge3\sqrt[3]{\dfrac{a^9.\left(b^3+c^2\right)}{8\left(b^3+c^2\right)}}=\dfrac{3a^3}{2}\)
Tương tự và cộng lại:
\(\Rightarrow M-\dfrac{a^4+b^4+c^4}{2}+\dfrac{a^3+b^3+c^3}{4}+\dfrac{a^2+b^2+c^2}{4}\ge\dfrac{3}{2}\left(a^3+b^3+c^3\right)\)
\(\Rightarrow M\ge\dfrac{a^4+b^4+c^4}{2}+\dfrac{5}{4}\left(a^3+b^3+c^3\right)-\dfrac{3}{4}\)
Mặt khác ta có:
\(\dfrac{1}{2}\left(a^4+b^4+c^4\right)\ge\dfrac{1}{6}\left(a^2+b^2+c^2\right)^2=\dfrac{3}{2}\)
\(\left(a^3+a^3+1\right)+\left(b^3+b^3+1\right)+\left(c^3+c^3+1\right)\ge3\left(a^2+b^2+c^2\right)=9\)
\(\Rightarrow2\left(a^3+b^3+c^3\right)+3\ge9\Rightarrow a^3+b^3+c^3\ge3\)
\(\Rightarrow M\ge\dfrac{3}{2}+\dfrac{15}{4}-\dfrac{3}{4}=...\)
6 tháng 5 2017
a)ĐKXĐ:\(a\ge0;a\ne16\)
\(B=\left[\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right]:\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
=\(\dfrac{3\sqrt{a}\left(\sqrt{a}-4\right)+\sqrt{a}\left(\sqrt{a}+4\right)-4\left(a+2\right)}{a-16}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}=\dfrac{-8\sqrt{a}-8}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8\left(-\sqrt{a}-1\right)}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8}{\sqrt{a}-4}\)
Vậy...
b)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
B=-3 thì \(\dfrac{8}{\sqrt{a}-4}=-3\)
=>\(9=-3\sqrt{a}+24\)
<=>-15=-3\(\sqrt{a}\)
<=>\(\sqrt{a}=5\)
<=>a=25(TM)
Vậy a=25 thì B=-3
c)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
Để B nguyên thì \(\dfrac{8}{\sqrt{a}-4}\)phải nguyên<=>8 chia hết cho \(\sqrt{a}-4\) <=>\(\sqrt{a}-4\)là Ư(8) Mà Ư(8)={-8;-4;-2;-1;1;2;4;8} Do \(\sqrt{a}\ge0\) ta có bảng sau:| \(\sqrt{a}-4\) | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| \(\sqrt{a}\) | -4(L) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| \(\sqrt{a}\) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| a | 0(TM) | 4(TM) | 9(TM) | 25(TM) | 36(TM) | 64(TM) | 144(TM) |
(BẠN KẺ 1 BẢNG 3 HÀNG THÔI NHA,MÌNH KẺ LỖI NÊN LÀM 2 BẢNG)
Vậy...
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 7 2021
Đặt vế trái là P
Ta có:
\(\dfrac{a}{b^3+ab}=\dfrac{a}{b\left(a+b^2\right)}=\dfrac{1}{b}-\dfrac{b}{a+b^2}\ge\dfrac{1}{b}-\dfrac{b}{2\sqrt{ab^2}}=\dfrac{1}{b}-\dfrac{1}{2\sqrt{a}}\ge\dfrac{1}{b}-\dfrac{1}{4}\left(\dfrac{1}{a}+1\right)\)
Tương tự và cộng lại:
\(P\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)
\(P\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}-\dfrac{3}{4}=\dfrac{3}{2}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
27 tháng 1 2022
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).64}}=\dfrac{3x}{4}\)
\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)
\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{x+y+z}{2}-\dfrac{3}{4}\ge\dfrac{3\sqrt[3]{xyz}}{2}-\dfrac{3}{4}=\dfrac{3}{2}-\dfrac{3}{4}=\dfrac{3}{4}\left(đpcm\right)\)
(bài này chắc thiếu đk xyz=1 ?nên mình bổ sung xyz=1)
OP
27 tháng 1 2022
( xyz=3)
Áp dụng BDDT AM-GM:
Ta có: \(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{\left(1+y\right)\left(1+z\right).8.8}}=3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)
Chứng minh tương tự ta có:
\(\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\)
\(\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{8}+\dfrac{1+y}{8}\ge\dfrac{3z}{4}\)
Cộng từng vế ta được:
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}+\dfrac{3+x+y+z}{4}\ge\dfrac{3\left(x+y+z\right)}{4}\)
\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{3x+3y+3z-3-x-y-z}{4}=\dfrac{2\left(x+y+z\right)-3}{4}\)
\(\Leftrightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+x\right)\left(1+z\right)}+\dfrac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\dfrac{2.\sqrt[3]{xyz}-3}{4}=\dfrac{2.3-3}{4}=\dfrac{3}{4}\left(đfcm\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 8 2021
Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)
BĐT trở thành:
\(\dfrac{y^2}{xz}+\dfrac{z^2}{xy}+\dfrac{x^2}{yz}\ge\dfrac{3}{2}\left(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}-1\right)\)
\(\Leftrightarrow2\left(x^3+y^3+z^3\right)+3xyz\ge3x^2y+3y^2z+3z^2x\)
Áp dụng BĐT Schur ta có:
\(x^3+y^3+z^3+3xyz\ge x^2y+y^2z+z^2x+xy^2+yz^2+zx^2\)
\(\Rightarrow VT\ge\left(x^3+xy^2\right)+\left(y^3+yz^2\right)+\left(z^3+zx^2\right)+x^2y+y^2z+z^2x\ge3\left(x^2y+y^2z+z^2x\right)\)
NT
2
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1
Áp dụng BĐT Holder:
\(\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\ge\left(a^2+b^2+c^2\right)^3\)
Mặt khác:
\(\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\dfrac{3}{2}\left(a^2b^2+b^2c^2+c^2a^2+abc\left(a+b+c\right)\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{3}{4}\left[a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\right]\)
\(\Rightarrow\left(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\right)^2\ge\dfrac{3}{4}\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}\)
\(\Rightarrow P\ge\dfrac{\sqrt{3}}{2}\sqrt{a^2+b^2+c^2}+\dfrac{4}{\sqrt{a^2+b^2+c^2+1}}\)
Đặt \(\sqrt{\dfrac{a^2+b^2+c^2}{3}}=x>0\)
\(\Rightarrow P\ge\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\)
Ta sẽ chứng minh \(P\ge\dfrac{7}{2}\)
Thật vậy, với \(x\ge\dfrac{7}{3}\Rightarrow P>\dfrac{3x}{2}\ge\dfrac{7}{2}\) (đúng)
Với \(0< x\le\dfrac{7}{3}\) ta cần chứng minh:
\(\dfrac{3x}{2}+\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7}{2}\Leftrightarrow\dfrac{4}{\sqrt{3x^2+1}}\ge\dfrac{7-3x}{2}\)
\(\Leftrightarrow64\ge\left(7-3x\right)^2\left(3x^2+1\right)\)
\(\Leftrightarrow3\left(x-1\right)^2\left(-9x^2+24x+5\right)\ge0\)
\(\Leftrightarrow\left(x-1\right)^2\left[3x\left(7-3x\right)+3x+5\right]\ge0\) (đúng)
Vậy \(P_{min}=\dfrac{7}{2}\) khi \(x=1\) hay \(a=b=c=1\)
\(a^4+b^4+c^4\ge a^3+b^3+c^3\)
\(\Leftrightarrow a^4+b^4+c^4-\left(a^3+b^3+c^3\right)\ge0\)
\(\Leftrightarrow a^4+b^4+c^4-\left(a^3+b^3+c^3\right)+3-\left(a+b+c\right)\ge0\)
\(\Leftrightarrow a^4-a^3-a+1+b^4-b^3-b+1+c^4-c^3-c+1\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(a^2+a+1\right)+\left(b-1\right)^2\left(b^2+b+1\right)+\left(c-1\right)^2\left(c^2+c+1\right)\ge0\) đúng với mọi a, b, c.
\(\Rightarrowđpcm\)
mong giúp mk bài này nữa vs ạ
https://hoc24.vn/cau-hoi/cho-ab-8-va-bge3-cm-27a210b3-945.1638861610920