Cho \(a+b+c=0\) và \(a,b,c\ne0\)
Chứng minh:
\(A=\sqrt{\dfrac{6a^2}{a^2-b^2-c^2}+\dfrac{6b^2}{b^2-c^2-a^2}+\dfrac{6c^2}{c^2-a^2-b^2}}\) là số nguyên
Có ai giỏi toán không
Cho a+b+c=0 và a,b,c\(\ne0\) . Chứng minh rằng:
A=\(\sqrt{\dfrac{6a^2}{a^2-b^2-c^2}+\dfrac{6b^2}{b^2-c^2-a^2}+\dfrac{6c^2}{c^2-a^2-b^2}}\) là số nguyên
Ta có:
\(a^2=\left(-b-c\right)^2\)
\(\Leftrightarrow a^2-b^2-c^2=2bc\)
Tương tự ta cũng có
\(\left\{{}\begin{matrix}b^2-c^2-a^2=2ca\\c^2-a^2-b^2=2ab\end{matrix}\right.\)
Thế vô ta được
\(A=\sqrt{\dfrac{3a^2}{bc}+\dfrac{3b^2}{ca}+\dfrac{3c^2}{ab}}\)
\(=\sqrt{\dfrac{3\left(a^3+b^3+c^3\right)}{abc}}\)
\(=\sqrt{3.\dfrac{\left(a^3+b^3+c^3-3abc\right)+3abC}{abc}}\)
\(=\sqrt{3.\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{abc}}\)
\(=\sqrt{3.3}=3\)
ĐPCM
1)Cho a;b;c>0 thỏa \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=4\)
Chứng minh \(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le1\)
2) Cho a;b;c>0
CMR \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Cho a;b;c>0 thỏa a+b+c=3
CMR \(\dfrac{a+b}{\sqrt{a^2+b^2+6c}}+\dfrac{b+c}{\sqrt{b^2+c^2+6a}}+\dfrac{c+a}{\sqrt{c^2+a^2+6b}}>2\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
Cho a,b,c>0 CMR
\(\sqrt{\dfrac{a^2}{6a^2+5ab+b^2}}+\sqrt{\dfrac{b^2}{6b^2+5bc+c^2}}+\sqrt{\dfrac{c^2}{6c^2+5ca+a^2}}\le\dfrac{\sqrt{3}}{2}\)
Cho a, b, c \(\ne0\) và a+b+c=0. CMR :
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\) là số hữu tỉ
Ta có: \(2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{2\left(a+b+c\right)}{abc}=0\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) là số hữu tỉ
Cho \(a,b,c\) là các số tự nhiên khác \(0\), \(a\ne c\) sao cho \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\). Chứng minh rằng \(a^2+b^2+c^2\) không phải là số nguyên tố.
cho a,b,c >0, và \(a^2+b^2+c^2=3\):
CMR: \(\dfrac{a^2+3ab+b^2}{\sqrt{6a^2+8ba+11b^2}}+\dfrac{a^2+3ab+c^2}{\sqrt{6a^2+8ca+11c^2}}+\dfrac{c^2+3cb+b^2}{\sqrt{6c^2+8ca+11b^2}}\) \(\leq\) 3
Lời giải:
Đặt biểu thức đã cho là \(A\)
Ta có:
\(6a^2+8ab+11b^2=2a^2+(2a+2b)^2+7b^2\)
Áp dụng BĐT Bunhiacopxky:
\([2a^2+(2a+2b)^2+7b^2](2+4^2+7)\geq (2a+8a+8b+7b)^2\)
\(\Leftrightarrow 25(6a^2+8ab+11b^2)\geq (10a+15b)^2\)
\(\Rightarrow \sqrt{6a^2+8ab+11b^2}\geq 2a+3b\)
\(\Rightarrow \frac{a^2+3ab+b^2}{\sqrt{6a^2+8ab+11b^2}}\leq \frac{a^2+3ab+b^2}{2a+3b}\)
Thực hiện tương tự với các biểu thức còn lại và cộng theo vế:
\(A\leq \frac{a^2+3ab+b^2}{2a+3b}+\frac{a^2+3ac+c^2}{2c+3a}+\frac{b^2+3bc+c^2}{2b+3c}\)
\(6A\leq \frac{3a(2a+3b)+2b(2a+3b)+5ab}{2a+3b}+\frac{3c(2c+3a)+2a(2c+3a)+5ac}{2c+3a}+\frac{3b(2b+3c)+2c(2b+3c)+5bc}{2b+3c}\)
\(\Leftrightarrow 6A\leq 3a+2b+\frac{5ab}{2a+3b}+3c+2a+\frac{5ac}{2c+3a}+3b+2c+\frac{5bc}{2b+3c}\)
\(\Leftrightarrow 6A\leq 5(a+b+c)+5\left(\frac{ab}{2a+3b}+\frac{bc}{2b+3c}+\frac{ac}{2c+3a}\right)\)
Theo hệ quả của BĐT AM-GM:
\((a+b+c)^2\leq 3(a^2+b^2+c^2)=9\Rightarrow a+b+c\leq 3(1)\)
Áp dụng BĐT Cauchy-Schwarz dạng ngược:
\(\frac{ab}{2a+3b}\leq \frac{ab}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}\right)\)
\(\frac{bc}{2b+3c}\leq \frac{bc}{25}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{ca}{2c+3a}\leq \frac{ca}{25}\left(\frac{1}{c}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}+\frac{1}{a}\right)\)
\(\Rightarrow \frac{ab}{2a+3b}+\frac{bc}{2b+3c}+\frac{ac}{2c+3a}\leq \frac{1}{5}(a+b+c)(2)\)
Từ (1); (2) suy ra:
\(6A\leq 5(a+b+c)+5.\frac{1}{5}(a+b+c)=6(a+b+c)\leq 18\)
\(\Rightarrow A\leq 3\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)
Cho \(a+b+c=a^2+b^2+c^2=1\) và \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) \(\left(a\ne0,b\ne0,c\ne0\right)\)
Chứng minh rằng: \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
Lời giải:
Đặt $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=t$
$\Rightarrow x=at; y=bt; z=ct$. Ta có:
$(x+y+z)^2=(at+bt+ct)^2=t^2(a+b+c)^2=t^2(*)$
Mặt khác:
$x^2+y^2+z^2=(at)^2+(bt)^2+(ct)^2=t^2(a^2+b^2+c^2)=t^2(**)$
Từ $(*); (**)\Rightarrow (x+y+z)^2=x^2+y^2+z^2$ (đpcm)
Cho \(a+b+c=0;a,b,c\ne0\)
Chứng minh hằng đẳng thức:
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)
Ta có:
\(VT=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}\)
\(=\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)-2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{c}{abc}+\dfrac{a}{abc}+\dfrac{b}{bca}\right)}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2-2\left(\dfrac{a+b+c}{abc}\right)}\)
\(=\sqrt{\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\)
\(=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\)
\(\Rightarrow VT=VP\)
Vậy \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}}=\left|\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right|\) (Đpcm)
Cho a,b,c là các số thực không âm thỏa mãn a+b+c = 1011. Chứng minh rằng:
\(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\) + \(\sqrt{2022b+\dfrac{\left(c-a\right)^2}{2}}\)+\(\sqrt{2022c+\dfrac{\left(a-b\right)^2}{2}}\) ≤ \(2022\sqrt{2}\)
Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\)
\(=\sqrt{2a\left(a+b+c\right)+\dfrac{b^2-2bc+c^2}{2}}\)
\(=\sqrt{\dfrac{4a^2+b^2+c^2+4ab+4ac-2bc}{2}}\)
\(=\sqrt{\dfrac{\left(2a+b+c\right)^2-4bc}{2}}\)
\(\le\sqrt{\dfrac{\left(2a+b+c\right)^2}{2}}\)
\(=\dfrac{2a+b+c}{\sqrt{2}}\).
Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)
\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)
ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.