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\(\left\{{}\begin{matrix}\dfrac{a+b}{6}=\dfrac{b+c}{5}\\\dfrac{a+b}{6}=\dfrac{c+a}{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{a}{2}\\c=\dfrac{3a}{4}\end{matrix}\right.\)
\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\dfrac{a^2}{4}+\dfrac{9a^2}{16}-a^2}{2.\dfrac{a}{2}.\dfrac{3a}{4}}=-\dfrac{1}{4}\)
\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{a^2+\dfrac{9a^2}{16}-\dfrac{a^2}{4}}{2a.\dfrac{3a}{4}}=\dfrac{7}{8}\)
\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{11}{16}\)
\(P=-\dfrac{1}{4}+\dfrac{14}{8}+\dfrac{44}{16}=\dfrac{17}{4}\)
Ta có : \(P=a^2+b^2+c^2\)
\(\Rightarrow P+2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)
\(\Rightarrow P+2=\left(a+b+c\right)^2\ge0\)
\(\Rightarrow P\ge-2\)
Vậy MinP = -2 tại a + b + c = 0 .
Mik thấy a,b,c>0 \(\Rightarrow a+b+c>0\)
\(\Rightarrow2P-2=2a^2+2b^2+2c^2-2ab-2bc-2ca=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) \(\Rightarrow2P\ge2\Rightarrow P\ge1\) Dấu bằng xảy ra \(\Leftrightarrow a=b=c=\dfrac{\sqrt{3}}{3}\) Vậy...
\(\left(a+b+c\right)\left(a+b-c\right)=3ab\)
\(\Leftrightarrow\left(a+b\right)^2-c^2=3ab\)
\(\Leftrightarrow a^2+b^2+2ab-c^2=3ab\)
\(\Leftrightarrow a^2+b^2-c^2=ab\)
\(\Leftrightarrow\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{2}\)
\(\Rightarrow cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{2}\)
\(\Rightarrow C=60^0\)
\(\left(a^2+b^2+c^2+1\right)x=ab+bc+ca\)
\(\Leftrightarrow x=\dfrac{ab+bc+ca}{a^2+b^2+c^2+1}\)
Ta có:
\(x^2-1=\dfrac{\left(ab+bc+ca\right)^2}{\left(a^2+b^2+c^2+1\right)^2}-1=\dfrac{\left(ab+bc+ca-a^2-b^2-c^2-1\right)\left(ab+bc+ca+a^2+b^2+c^2+1\right)}{\left(a^2+b^2+c^2+1\right)^2}\)
\(=\dfrac{\left[-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2-2\right]\left[\left(a+b+c\right)^2+a^2+b^2+c^2+2\right]}{4\left(a^2+b^2+c^2+1\right)^2}< 0\)
\(\Rightarrow x^2-1< 0\Rightarrow\left|x\right|< 1\)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(T=\frac{\frac{1}{a^2}}{\frac{1}{b}+\frac{1}{c}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}=\frac{1}{2}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)
\(\geq \frac{1}{2}.3\sqrt[3]{\frac{1}{abc}}=\frac{3}{2}\) (theo BĐT AM-GM)
Vậy $T_{\min}=\frac{3}{2}$.
Giá trị này đạt tại $a=b=c=1$
Áp dụng Cauchy-Schwarz, ta có:
\(VT\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)
\(VT\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)
Chọn C.
Ta có: AB + BC = AC nên ba điểm A; B; C thẳng hàng và B nằm giữa A; C
Khi đó