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a) CM:\(\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
\(\Leftrightarrow n+1+n=\left(n+1-n\right)\left(n+1+n\right)\)
\(\Leftrightarrow2n+1=1\left(2n+1\right)\)
\(\Leftrightarrow2n+1=2n+1\)
\(\Rightarrow\sqrt{\left(n+1\right)^2}+\sqrt{n^2}=\left(n+1\right)^2-n^2\)
Câu b) ý 2:
Áp dụng BĐT cô si ta có :
\(\dfrac{a}{b}+\dfrac{b}{c}\ge2\sqrt{\dfrac{a}{c}}\\ \dfrac{b}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{b}{a}}\\ \dfrac{c}{a}+\dfrac{a}{b}\ge2\sqrt{\dfrac{c}{b}}\\ \Leftrightarrow2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{b}{a}}+\sqrt{\dfrac{c}{b}}\right)\\ \Rightarrowđpcm\)
Lời giải:
Từ \(\sqrt{a}+\sqrt{b}=1\Rightarrow (\sqrt{a}+\sqrt{b})^2=1\)
\(\Rightarrow a+b+2\sqrt{ab}=1\)
Áp dụng BĐT Cô-si cho các số dương:
\(1=(a+b)+2\sqrt{ab}\geq 2\sqrt{(a+b).2\sqrt{ab}}\)
\(\Rightarrow 1\geq 4(a+b).2\sqrt{ab}\) (bình phương 2 vế)
\(\Rightarrow \frac{1}{8}\geq (a+b)\sqrt{ab}\)
Ta cớ đpcm
Dấu "=" xảy ra khi \(a=b=\frac{1}{4}\)
Bài 1:
a: \(=\dfrac{1}{mn^2}\cdot\dfrac{n^2\cdot\left(-m\right)}{\sqrt{5}}=\dfrac{-\sqrt{5}}{5}\)
b: \(=\dfrac{m^2}{\left|2m-3\right|}=\dfrac{m^2}{3-2m}\)
c: \(=\left(\sqrt{a}+1\right):\dfrac{\left(a-1\right)^2}{\left(1-\sqrt{a}\right)}=\dfrac{-\left(a-1\right)}{\left(a-1\right)^2}=\dfrac{-1}{a-1}\)
Bài 1:
Với $a=0$ hoặc $b=0$ thì ta luôn có \(ab=a^ab^b\)
Với $a\neq 0; b\neq 0$ , tức là \(a,b\in (0;1]\)
Ta có: \(a^a-a=a(a^{a-1}-1)=a(\frac{1}{a^{1-a}}-1)=\frac{a}{a^{1-a}}(1-a^{1-a})\)
Với \(0\leq a\leq 1; 1-a\geq 0\Rightarrow a^{1-a}\leq 1\)
\(\Rightarrow 1-a^{1-a}\geq 0\)
\(\Rightarrow a^a-a=\frac{a}{a^{1-a}}(1-a^{1-a})\geq 0\)
\(\Rightarrow a^a\geq a\)
Tương tự: \(b^b\geq b\)
\(\Rightarrow a^ab^b\geq ab\) (đpcm)
Bài 2:
Ta có :\(\frac{1}{3^a}+\frac{1}{3^b}+\frac{1}{3^c}\geq 3\left(\frac{a}{3^a}+\frac{b}{3^b}+\frac{c}{3^c}\right)\)
\(\Leftrightarrow \frac{1-3a}{3^a}+\frac{1-3b}{3^b}+\frac{1-3c}{3^c}\geq 0\)
\(\Leftrightarrow \frac{b+c-2a}{3^a}+\frac{a+c-2b}{3^b}+\frac{a+b-2c}{3^c}\geq 0\) (do $a+b+c=1$)
\(\Leftrightarrow (a-b)\left(\frac{1}{3^b}-\frac{1}{3^a}\right)+(b-c)\left(\frac{1}{3^c}-\frac{1}{3^b}\right)+(c-a)\left(\frac{1}{3^a}-\frac{1}{3^c}\right)\geq 0\)
\(\Leftrightarrow \frac{(a-b)(3^a-3^b)}{3^{a+b}}+\frac{(b-c)(3^b-3^c)}{3^{b+c}}+\frac{(c-a)(3^c-3^a)}{3^{c+a}}\geq 0(*)\)
Ta thấy, với mọi \(a\geq b\Rightarrow 3^a\geq 3^b; a\leq b\Rightarrow 3^a\leq 3^b\)
Tức là \(a-b; 3^a-3^b\) luôn cùng dấu
\(\Rightarrow (a-b)(3^a-3^b)\geq 0\). Kết hợp với \(3^{a+b}>0, \forall a,b\)
\(\Rightarrow \frac{(a-b)(3^a-3^b)}{3^{a+b}}\geq 0\)
Tương tự: \(\frac{(b-c)(3^b-3^c)}{3^{b+c}}\geq 0; \frac{(c-a)(3^c-3^a)}{3^{c+a}}\geq 0\)
Do đó $(*)$ đúng, ta có đpcm.
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
a/ \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}=1+\dfrac{1}{2.2}+...+\dfrac{1}{n.n}\)
\(< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\)
\(=1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(=1+1-\dfrac{1}{n}=2-\dfrac{1}{n}< 2\)
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=1\)
\(\Rightarrow a+1=a+\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\)
Tương tự: \(b+1=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(c+1=\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)\)
\(VT=\sum\dfrac{\sqrt{a}}{a+1}=\sum\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\dfrac{2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(VP=\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}=\dfrac{2}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2}}\)
\(=\dfrac{2}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{b}+\sqrt{c}\right)}\)
\(\Rightarrow VT=VP\) (đpcm)
Use Be-loli 's ineq:
\(\left(\dfrac{2a}{a+b}\right)^n=\left(1+\dfrac{a-b}{a+b}\right)^n\ge1+\dfrac{n\left(a-b\right)}{a+b}\)
\(\left(\dfrac{2b}{a+b}\right)^n=\left(1-\dfrac{a-b}{a+b}\right)^n\ge1-\dfrac{n\left(a-b\right)}{a+b}\)
Cộng theo vế 2 BĐT trên ta có:
\(\left(\dfrac{2a}{a+b}\right)^n+\left(\dfrac{2b}{a+b}\right)^n\ge2\Leftrightarrow\left(\dfrac{a+b}{2}\right)^n\le\dfrac{a^n+b^n}{2}\)