Bai 1: Cho a2=b.c. Chứng minh: \(\dfrac{a^2+c^2}{a^2+b^2}=\dfrac{c}{d}\)
Bài 2: Cho: \(^{\dfrac{a}{b}=\dfrac{b}{c}}\). Chứng minh: \(\dfrac{a^2-b^2}{b^2-c^2}\)=\(\dfrac{a}{c}=\dfrac{\left(a-b\right)^2}{\left(b-c\right)^2}\)
GIÚP KHẨN CẤP NHA !
Bài 1: a;b;c > 0
Chứng minh : \(\dfrac{a}{3a+b+c}+\dfrac{b}{3b+a+c}+\dfrac{c}{3c+a+b}\le\dfrac{3}{5}\)
Bài 2: x;y;z \(\ne\) 1 và xyz = 1
Chứng minh : \(\dfrac{x^2}{\left(x-1\right)^2}+\dfrac{y^2}{\left(y-1\right)^2}+\dfrac{z^2}{\left(z-1\right)^2}\ge1\)
1.
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)
Tương tự:
\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)
Cộng vế:
\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)
Dấu "=" xảy ra khi \(a=b=c\)
2.
Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)
Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)
Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)
Biến đổi giả thiết:
\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)
\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
\(\Rightarrow ab+bc+ca=a+b+c-1\)
BĐT cần chứng minh trở thành:
\(a^2+b^2+c^2\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)
\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)
\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)
Cho các số thực a,b,c,d,e thỏa mãn \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)chứng minh rằng: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)=\dfrac{a^2}{b.c}\)
Sửa: CMR: \(\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3=\dfrac{a^2}{bc}\)
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\\ \Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2019b+2020c-2021d}{2019c+2020d-2021e}\right)^3\left(1\right)\\ \dfrac{a}{b}=\dfrac{b}{c}=k\Rightarrow a=bk;b=ck\Rightarrow a=ck^2\\ \Rightarrow\dfrac{a^2}{bc}=\dfrac{c^2k^4}{ck\cdot c}=k^3=\left(\dfrac{a}{b}\right)^3\left(2\right)\\ \left(1\right)\left(2\right)\RightarrowĐpcm\)
1) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) . Chứng minh rằng \(\dfrac{2a^2-3ab+5b^2}{2a^2+3ab}=\dfrac{2c^2-3cd+5d^2}{2c^2+3cd}\)
2) Cho \(\dfrac{a}{c}=\dfrac{c}{b}\). Chứng minh rằng \(\dfrac{b^2-c^2}{a^2+c^2}=\dfrac{b-a}{a}\)
3) Cho \(\dfrac{a}{b}=\dfrac{c}{d}\).Chứng minh rằng\(\dfrac{3a^6+c^6}{3b^6+d^6}=\dfrac{\left(a+c\right)^6}{\left(b+d\right)^6}\)
Bài 1:
$\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$. Khi đó:
\(\frac{2a^2-3ab+5b^2}{2a^2+3ab}=\frac{2(bt)^2-3.bt.b+5b^2}{2(bt)^2+3bt.b}=\frac{b^2(2t^2-3t+5)}{b^2(2t^2+3t)}\)
$=\frac{2t^2-3t+5}{2t^2+3t}(1)$
\(\frac{2c^2-3cd+5d^2}{2c^2+3cd}=\frac{2(dt)^2-3.dt.d+5d^2}{2(dt)^2+3dt.d}=\frac{d^2(2t^2-3t+5)}{d^2(2t^2+3t)}=\frac{2t^2-3t+5}{2t^2+3t}(2)\)
Từ $(1);(2)$ suy ra đpcm.
Bài 2:
Từ $\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab$. Khi đó:
$\frac{b^2-c^2}{a^2+c^2}=\frac{b^2-ab}{a^2+ab}=\frac{b(b-a)}{a(a+b)}$ (đpcm)
Bài 3:
Đặt $\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt$
Khi đó:
$\frac{3a^6+c^6}{3b^6+d^6}=\frac{3(bt)^6+(dt)^6}{3b^6+d^6}=\frac{t^6(3b^6+d^6)}{3b^6+d^6}=t^6(*)$
Và:
$\frac{(a+c)^6}{(b+d)^6}=(\frac{bt+dt}{b+d})^6=t^6(**)$
Từ $(*); (**)\Rightarrow $ đpcm.
1. Cho \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\). Chứng minh rằng \(\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\)
2. Cho \(\dfrac{a}{2003}=\dfrac{b}{2004}=\dfrac{c}{2005}\). Chứng minh rằng \(4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Bài 1:
Áp dụng t.c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)
Cho tỉ lệ thức \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng \(\dfrac{a.d}{c.d}=\dfrac{a^2-b^2}{b^2-d^2}\)và \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Đẳng thức đầu tiên sai:
Ví dụ: \(a=1;b=2;c=3;d=6\) thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Nhưng \(\dfrac{a.d}{c.d}\ne\dfrac{a^2-b^2}{b^2-d^2}\)
Với đẳng thức thứ 2:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
Chứng minh :
a, \(\dfrac{a+b+c}{3}\dfrac{>}{ }\sqrt{\dfrac{ab+bc+ca}{3}}\) với a,b,c>0
b,\(\dfrac{a^2+b^2+c^2}{3}\dfrac{>}{ }\left(\dfrac{a+b+c}{3}\right)^2\)
c,\(\dfrac{x^2+2}{\sqrt{x^2+1}}\dfrac{>}{ }2\)
d,\(\dfrac{a^3+b^3}{2}\dfrac{>}{ }\left(\dfrac{a+b}{2}\right)^3\)
a) Ta có:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^2}{9}\ge\dfrac{\left(ab+bc+ca\right)}{3}\)
\(\Leftrightarrow\dfrac{a+b+c}{3}\ge\sqrt{\dfrac{ab+bc+ca}{3}}\)
Đẳng thức xảy ra khi $a=b=c.$
b) BĐT \(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
Hay là \(2\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\),
đúng.
Đẳng thức xảy ra khi $a=b=c.$
c) \(\Leftrightarrow\dfrac{\left(x^2+2\right)^2}{x^2+1}\ge4\Leftrightarrow x^4+4x^2+4\ge4x^2+4\Leftrightarrow x^4\ge0\)
Đẳng thức xảy ra khi $x=0.$
d) Xét hiệu hai vế đi bạn.
Chứng minh:
a, \(a^3+b^3+c^3\dfrac{>}{ }3abc\)
b,\(abc\dfrac{< }{ }\left(\dfrac{a+b+c}{3}\right)^3\)
c,\(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\dfrac{< }{ }a+b+c\)
d,\(\dfrac{a}{b+c}+\dfrac{c}{a+b}+\dfrac{b}{a+c}\dfrac{>}{ }\dfrac{3}{2}\left(a,b,c>0\right)\)
Cho: \(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}=0\). Chứng minh: \(\dfrac{a}{\left(b-c\right)^2}+\dfrac{b}{\left(c-a\right)^2}+\dfrac{c}{\left(a-b\right)^2}=0\) trong đó a, b, c đôi 1 khác nhau và khác 0
Cho a;b;c là 3 cạnh tam giác. CHứng minh: \(\dfrac{1}{\left(a+b-c\right)^2}+\dfrac{1}{\left(a-b+c\right)^2}+\dfrac{1}{\left(-a+b+c\right)^2}\ge\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)
Lời giải:
Đặt \(\left\{\begin{matrix} a+b-c=x\\ b+c-a=y\\ c+a-b=z\end{matrix}\right.\Rightarrow \left\{\begin{matrix} b=\frac{x+y}{2}\\ c=\frac{y+z}{2}\\ a=\frac{x+z}{2}\end{matrix}\right.\) \((x,y,z>0\) do $a,b,c$ là ba cạnh tam giác ).
BĐT cần chứng minh tương đương với :
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)
Áp dụng BĐT Cauchy:
\(\frac{1}{x^2}+\frac{1}{y^2}\geq \frac{2}{xy}\)
\(\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \left(\frac{1}{x}+\frac{1}{y}\right)^2\)
Theo BĐT S.Vacso: \(\frac{1}{x}+\frac{1}{y}\geq \frac{4}{x+y}\Rightarrow 2\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq \frac{16}{(x+y)^2}(*)\)
Hoàn toàn tương tự:
\(2\left(\frac{1}{y^2}+\frac{1}{z^2}\right)\geq \frac{16}{(y+z)^2}; 2\left(\frac{1}{z^2}+\frac{1}{x^2}\right)\geq \frac{16}{(z+x)^2}(**)\)
Cộng theo vế \((*); (**)\) và rút gọn suy ra:
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{4}{(x+y)^2}+\frac{4}{(y+z)^2}+\frac{4}{(z+x)^2}\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z$ hay $a=b=c$
Cho a, b, c > 0 và abc = 1. Chứng minh rằng \(\dfrac{1}{a^2.\left(b+c\right)}+\dfrac{1}{b^2.\left(c+a\right)}+\dfrac{1}{c^2.\left(a+b\right)}\ge\dfrac{3}{2}\)
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.