Cho tam giác ABC nhọn có BC = a, AC = b và AB = C. Chứng minh:
a) \(\dfrac{a}{sin_A}=\dfrac{b}{sin_B}=\dfrac{c}{sin_C}\)
b) \(\sqrt{a.sin_A}+\sqrt{b.sin_B}+\sqrt{c.sin_C}=\sqrt{\left(a+b+c\right)\left(sin_A+sin_B+sin_C\right)}\)
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Dễ dàng c/m : \(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}=1\)
Ta có : \(\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le\dfrac{1}{a+b+4}\le\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}\right)\)
Suy ra : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)
" = " \(\Leftrightarrow a=b=c=1\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$
$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$
Cộng theo vế và thu gọn:
$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$
$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$
Ta có đpcm.
Bài 2:
$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$
$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$
$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$
Cộng theo vế và rút gọn thu được:
$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b=c$
Vẽ AI, BI, CI cắt các cạnh đối diện thứ tự tại D,E,F.
Ta có công thức đường phân giác như sau:
\( AD^2 = \frac{{bc\left( {a + b + c} \right)\left( {b + c - a} \right)}}{{\left( {b + c} \right)^2 }} \)
Ta có:
\( \begin{array}{l} \frac{{IA}}{{ID}} = \frac{{BA}}{{BD}} = \frac{{CA}}{{CD}} = \frac{{b + c}}{a} \Leftrightarrow \frac{{IA}}{{AD}} = \frac{{b + c}}{{a + b + c}} \\ \Leftrightarrow IA^2 = AD^2 .\frac{{\left( {b + c} \right)^2 }}{{\left( {a + b + c} \right)^2 }} = \frac{{bc\left( {a + b + c} \right)\left( {b + c - a} \right)}}{{\left( {b + c} \right)^2 }}.\frac{{\left( {b + c} \right)^2 }}{{\left( {a + b + c} \right)^2 }} = \frac{{\left( {b + c - a} \right)bc}}{{a + b + c}} \\ \Leftrightarrow \frac{{IA^2 }}{{bc}} = \frac{{b + c - a}}{{a + b + c}} \\ \end{array} \)
Điều phải chứng minh
b) Từ câu a) ta suy ra được
\(\frac{IA^{^{2}}}{AB.AC}+\frac{IB^{2}}{BA.BC}+\frac{IC^{2}}{CA.CB}=1\)
\(\Leftrightarrow aIA^2+bIB^2+cIC^2=abc\)
Sử dụng BĐT Cauchy-Schwarz, ta có:
\(\left(IA+IB+IC\right)^2=\left(\dfrac{\sqrt{a}.IA}{\sqrt{a}}+\dfrac{\sqrt{b}.IB}{\sqrt{b}}+\dfrac{\sqrt{c}.IC}{\sqrt{c}}\right)^2\)
\(\le\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(aIA^2+bIB^2+cIC^2\right)\)
\(=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)abc=ab+bc+ca\)
\(\Rightarrow IA+IB+IC\le\sqrt{ab+bc+ca}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+ac+bc}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{a}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\) Chứng minh tương tự ta được:
\(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{b+a}+\dfrac{b}{b+c}\right);\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\)
\(\Rightarrow\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+a}+\dfrac{b}{b+c}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)=\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\left(1+1+1\right)=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{\sqrt{3}}\)
\(\dfrac{a}{\sqrt{a^2+1}}=\dfrac{a}{\sqrt{a^2+ab+bc+ca}}=\dfrac{a}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}\right)\)
Tương tự: \(\dfrac{b}{\sqrt{b^2+1}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{b}{b+c}\right)\) ; \(\dfrac{c}{\sqrt{c^2+1}}\le\dfrac{1}{2}\left(\dfrac{c}{c+a}+\dfrac{c}{b+c}\right)\)
Cộng vế:
\(VT\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{b}{a+b}+\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
Không ai thảo luận câu này sao. T khởi động trước nhá.
Ta có: \(\cos\left(\dfrac{A-B}{2}\right)=\dfrac{\cos\left(\dfrac{A-B}{2}\right).\cos\left(\dfrac{A+B}{2}\right)}{\sin\dfrac{C}{2}}\)
\(=\dfrac{\cos A+\cos B}{2\sqrt{\dfrac{1-\cos C}{2}}}=\dfrac{\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{a^2+c^2-b^2}{2ca}}{2\sqrt{\dfrac{1-\dfrac{a^2+b^2-c^2}{2ab}}{2}}}\)
\(=\dfrac{\dfrac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{abc}}{2\sqrt{\dfrac{c^2-\left(a-b\right)^2}{ab}}}=\dfrac{\left(a+b\right)\sqrt{c^2-\left(a-b\right)^2}}{2c\sqrt{ab}}\)
Ta sẽ chứng minh: \(\dfrac{\left(a+b\right)\sqrt{c^2-\left(a-b\right)^2}}{2c\sqrt{ab}}\le\dfrac{a+b}{\sqrt{2\left(a^2+b^2\right)}}\)
\(\Leftrightarrow\dfrac{2abc^2}{c^2-\left(a-b\right)^2}\ge a^2+b^2\)
\(\Leftrightarrow2abc^2-\left(a^2+b^2\right)\left(c^2-\left(a-b\right)^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2-c^2\right)\ge0\) (đúng vì tam giác ABC nhọn)
\(\Rightarrow\cos\left(\dfrac{A-B}{2}\right)\le\dfrac{a+b}{\sqrt{2\left(a^2+b^2\right)}}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\cos\left(\dfrac{B-C}{2}\right)\le\dfrac{b+c}{\sqrt{2\left(b^2+c^2\right)}}\left(2\right)\\\cos\left(\dfrac{C-A}{2}\right)\le\dfrac{c+a}{\sqrt{2\left(c^2+a^2\right)}}\left(3\right)\end{matrix}\right.\)
Cộng (1), (2), (3) vế theo vế ta được ĐPCM.
Thảo luận mình là người thứ 2
Chẳng thấy đề có kết nối giữa hai đại lượng [(ABC);(a,b,c)]
gì cả --> thiếu mối liên lạc cần thiết -->đề chưa thực sự rõ rằng --->Đề có suy biến --->lời giải (nếu có) phải là lời giải biện luận theo đề--->chưa thể chấp nhận lời giải trên
Xét \(\sqrt{\dfrac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}=\sqrt{\dfrac{\left(a\left(a+b+c\right)+bc\right)\left(b\left(a+b+c\right)+ac\right)}{c\left(a+b+c\right)+ab}}\)
\(=\sqrt{\dfrac{\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)}{ac+bc+c^2+ab}}\)
\(=\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}}\)\(=\sqrt{\left(a+b\right)^2}=a+b\)
Tương tự cho 2 đẳng thức còn lại rồi cộng theo vế
\(P=a+b+b+c+c+a=2\left(a+b+c\right)=2\)
Câu a: tự chứng minh.
Câu b: áp dụng câu a![Kết quả hình ảnh cho troll face full](data:image/png;base64,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)
câu a dùng định lí hàm sin(Trong SGK nhé bạn)