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\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd-b^2cd=abc^2+abd^2\)
\(\Leftrightarrow a^2cd-abc^2-abd^2+b^2cd=0\)
\(\Leftrightarrow ac\left(ad-bc\right)-bd\left(ad-bc\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}ac-bd=0\\ad-bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}ac=bd\\ad=bc\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{d}{c}\\\dfrac{a}{b}=\dfrac{c}{d}\end{matrix}\right.\) (ĐPCM)
1) Áp dụng bất đẳng Bunyakovsky dạng cộng mẫu ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}=\frac{a^6}{abc}+\frac{b^6}{abc}+\frac{c^6}{abc}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\)
\(=\frac{\left(a^3+b^3+c^3\right)\left(a^3+b^3+c^3\right)}{3abc}\ge\frac{3abc\left(a^3+b^3+c^3\right)}{3abc}=a^3+b^3+c^3\)
(Cauchy 3 số) Dấu "=" xảy ra khi: a = b = c
2) Áp dụng kết quả phần 1 ta có:
\(\frac{a^5}{bc}+\frac{b^5}{ca}+\frac{c^5}{ab}\ge\frac{\left(a^3+b^3+c^3\right)^2}{3abc}\ge\frac{\left(a^3+b^2+c^3\right)^2}{3\cdot\frac{1}{3}}=\left(a^3+b^3+c^3\right)^2\)
Dấu "=" xảy ra khi: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
https://hoc24.vn/hoi-dap/tim-kiem?q=Cho+c%C3%A1c+s%E1%BB%91+th%E1%BB%B1c+d%C6%B0%C6%A1ng+a,+b,+c+tho%E1%BA%A3+m%C3%A3n:+abc+a+b=3ababc+a+b=3ababc+a+b=3ab.+Ch%E1%BB%A9ng+minh+r%E1%BA%B1ng:+%E2%88%9Aaba+b+1+%E2%88%9Abbc+c+1+%E2%88%9Aaca+c+1%E2%89%A5%E2%88%9A3aba+b+1+bbc+c+1+aca+c+1%E2%89%A53\sqrt{\dfrac{ab}{a+b+1}}+\sqrt{\dfrac{b}{bc+c+1}}+\sqrt{\dfrac{a}{ca+c+1}}\ge\sqrt{3}&id=695796
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 \(P.Q=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc}.\dfrac{9abc}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=9\)
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 \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc\)
\(\Rightarrow sin\dfrac{A}{2}.sin\dfrac{B}{2}.sin\dfrac{C}{2}\le\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{1}{8}\)
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).
Áp dụng bất đẳng thức cô si ta có, với a,b,c >0
a/bc + b/ac ≥ 2*1/c
b/ac + c/ab ≥ 2*1/a
a/bc + c/ab ≥ 2*1/b
Cộng từng vế của 3 bất đẳng thức trên với nhau ta được
2*(a/bc + b/ac + c/ab) ≥ 2(1/a+1/b+1/c)
<=> đpcm
Sử dụng bất đẳng thức Cô si cho hai số dương ta có:
\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{b}{ca}}=\dfrac{2}{b}bca+cab≥2bca.cab=b2
Viết hai bất đẳng thức tương tự rồi cộng theo vế ba bất đẳng thức nhận được rồi chia 2 vế bất đẳng thức cho 2 ta được đpcm.
\(\dfrac{ab+1}{b}=\dfrac{bc+1}{c}=\dfrac{ca+1}{a}\)
\(\Leftrightarrow a+\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{1}{c}-\dfrac{1}{b}\\b-c=\dfrac{1}{a}-\dfrac{1}{c}\\c-a=\dfrac{1}{b}-\dfrac{1}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{b-c}{bc}\\b-c=\dfrac{c-a}{ca}\\c-a=\dfrac{a-b}{ab}\end{matrix}\right.\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{a^2b^2c^2}\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(1-\dfrac{1}{a^2b^2c^2}\right)=0\)
Dễ thấy \(1-\dfrac{1}{a^2b^2c^2}\ne0\left(abc\ne\pm1\right)\)
\(\Leftrightarrow a=b=c\) ( đpcm )