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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 $a=b=c=\frac{1}{3}$
Ta có \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\Leftrightarrow\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=0\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ac+c^2-3ab\right)=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2ac-2bc=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\left(tm\right)\\a=b=c\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a+b+c=0\)\(\Leftrightarrow\left[{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)
Ta có \(P=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\Leftrightarrow abc.P=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)=ab\left(a-b\right)-bc\left(a-b+c-a\right)+ca\left(c-a\right)=ab\left(a-b\right)-bc\left(a-b\right)-bc\left(c-a\right)+ca\left(c-a\right)=b\left(a-b\right)\left(a-c\right)-c\left(b-a\right)\left(c-a\right)=\left(a-b\right)\left(a-c\right)\left(b-c\right)\Leftrightarrow P=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc}\)\(Q=\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).Q=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)=c\left(b-c\right)\left(c-a\right)-\left(c+b\right)\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)=c\left(b-c\right)\left(c-a\right)-c\left(a-b\right)\left(c-a\right)-b\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)=c\left(c-a\right)\left(2b-c-a\right)-b\left(a-b\right)\left(2c-a-b\right)=c\left(c-a\right)3b-b\left(a-b\right)3c=3bc\left(b+c-2a\right)=-9abc\Leftrightarrow Q=\dfrac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{9abc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)Vậy \(P.Q=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc}.\dfrac{9abc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=9\)
Xét: \(\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\)
\(\Leftrightarrow a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}+1-\dfrac{a^2}{a^2+1}+1-\dfrac{b^2}{b^2+1}+1-\dfrac{c^2}{c^2+1}\)
\(\Leftrightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)+3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Xét \(3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy ta có
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\) ( 3 )
Xét \(3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2}{a^2+1}\le\dfrac{a^2}{2a}=\dfrac{a}{2}\\\dfrac{b^2}{b^2+1}\le\dfrac{b^2}{2b}=\dfrac{b}{2}\\\dfrac{c^2}{c^2+1}\le\dfrac{c^2}{2c}=\dfrac{c}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\le\dfrac{a+b+c}{2}=\dfrac{3}{2}\)
\(\Rightarrow3-\left(\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\right)\ge3-\dfrac{3}{2}=\dfrac{3}{2}\) ( 4 )
Từ ( 3 ) và ( 4 ) cộng theo từng vế
\(\Rightarrow VT\ge\dfrac{3}{2}+\dfrac{3}{2}=3\)
\(\Leftrightarrow\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge3\)
\(\Rightarrow\) ( đpcm )
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)
Từ giả thiết, ta có:
\(\dfrac{1}{1+a}\ge1-\dfrac{1}{1+b}+1-\dfrac{1}{1+c}+1-\dfrac{1}{1+d}=\dfrac{b}{1+b}+\dfrac{c}{c+1}+\dfrac{d}{d+1}\ge3\sqrt[3]{\dfrac{b.c.d}{\left(1+b\right)\left(1+c\right)\left(1+d\right)}}\)
Tương tự:
\(\dfrac{1}{1+b}\ge3\sqrt[3]{\dfrac{cda}{\left(1+a\right)\left(1+c\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+c}\ge3\sqrt[3]{\dfrac{abd}{\left(1+a\right)\left(1+b\right)\left(1+d\right)}}\)
\(\dfrac{1}{1+d}\ge3\sqrt[3]{\dfrac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Nhân vế theo vế 4 BĐT vừa chứng minh rồi rút gọn ta được:
\(abcd\le\dfrac{1}{81}\left(đpcm\right)\)
3)kẻ BD vuông góc voi71 BC, D thuộc AC
tam giác ABC cân tại A có AH là Đường cao
suy ra AH là trung tuyến
Suy ra BH=HC
(BD vuông góc BC
AH vuông góc BC
suy ra BD song song AH
suy ra BD/AH = BC/CH = 2
suyra 1/BD = 1/2AH suy ra 1BD^2 =1/4AH^2
tam giác BDC vuông tại B có BK là đường cao
suy ra 1/BK^2 =1/BD^2 +1/BC^2
suy ra 1/BK^2 =1/4AH^2 +1/BC^2
1) \(1+tan^2\alpha=1+\dfrac{sin^2\alpha}{cos^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\) (đpcm).
A B C D H K a, Vẽ phân giác AD của góc BAC
Kẻ BH\(\perp\)AD tại H ; CK\(\perp AD\) tại K
Dễ thấy \(sin\widehat{A_1}=sin\widehat{A_2}=sin\dfrac{A}{2}=\dfrac{BH}{AB}=\dfrac{CK}{AC}=\dfrac{BH+CK}{AB+AC}\le\)\(\le\dfrac{BD+CD}{b+c}=\dfrac{a}{b+c}\)
b, Tượng tự \(sin\dfrac{B}{2}\le\dfrac{b}{a+c};sin\dfrac{C}{2}\le\dfrac{c}{a+b}\)
Mặt khác \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)
\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)