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Ap dung bdt Cauchy-Schwarz dang Engel co:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{\left(1+1\right)^2}{p-a+p-b}=\dfrac{4}{2p-a-b}=\dfrac{4}{c}\)
Tuong tu: \(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a}\);
\(\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\)
Cong theo ve cac bdt tren ta co:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
=> Đpcm
Ta đi chứng minh BĐT : \(a^2+b^2+c^2\ge2\left(bc+ac-ab\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab-2bc-2ac\ge0\)
\(\Leftrightarrow\) \(\left(a+b-c\right)^2\ge0\) luôn đúng.
\(\Rightarrow2\left(bc+ac-ab\right)\le\dfrac{5}{3}\)
\(\Leftrightarrow bc+ac-ab\le\dfrac{5}{6}< 1\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}< \dfrac{1}{abc}\)
Câu 3. Dự đoán dấu "=" khi \(a=b=c=\frac{1}{\sqrt{3}}\)
Dùng phương pháp chọn điểm rơi thôi :)
LG
Áp dụng bđt Cô-si được \(a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow1\ge3\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{3}\ge\sqrt[3]{a^2b^2c^2}\)
\(\Rightarrow\frac{1}{27}\ge a^2b^2c^2\)
\(\Rightarrow\frac{1}{\sqrt{27}}\ge abc\)
Khi đó :\(B=a+b+c+\frac{1}{abc}\)
\(=a+b+c+\frac{1}{9abc}+\frac{8}{9abc}\)
\(\ge4\sqrt[4]{abc.\frac{1}{9abc}}+\frac{8}{9.\frac{1}{\sqrt{27}}}\)
\(=4\sqrt[4]{\frac{1}{9}}+\frac{8\sqrt{27}}{9}=\frac{4}{\sqrt[4]{9}}+\frac{8}{\sqrt{3}}=\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4\sqrt{3}\)
Dấu "=" \(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\)
Vậy .........
2, \(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)
\(A=\left[\frac{a^2}{b+c}+\frac{\left(b+c\right)}{4}\right]+\left[\frac{b^2}{a+c}+\frac{\left(a+c\right)}{4}\right]+\left[\frac{c^2}{a+b}+\frac{\left(a+b\right)}{4}\right]-\frac{\left(a+b+c\right)}{2}\)
Áp dụng BĐT AM-GM ta có:
\(A\ge2.\sqrt{\frac{a^2}{4}}+2.\sqrt{\frac{b^2}{4}}+2.\sqrt{\frac{c^2}{4}}-\frac{\left(a+b+c\right)}{2}\)
\(A\ge a+b+c-\frac{6}{2}\)
\(A\ge6-3\)
\(A\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow\)\(\frac{a^2}{b+c}=\frac{b+c}{4}\Leftrightarrow4a^2=\left(b+c\right)^2\Leftrightarrow2a=b+c\)(1)
\(\frac{b^2}{a+c}=\frac{a+c}{4}\Leftrightarrow4b^2=\left(a+c\right)^2\Leftrightarrow2b=a+c\)(2)
\(\frac{c^2}{a+b}=\frac{a+b}{4}\Leftrightarrow4c^2=\left(a+b\right)^2\Leftrightarrow2c=a+b\)(3)
Lấy \(\left(1\right)-\left(3\right)\)ta có:
\(2a-2c=c+b-a-b=c-a\)
\(\Rightarrow2a-2c-c+a=0\)
\(\Leftrightarrow3.\left(a-c\right)=0\)
\(\Leftrightarrow a-c=0\Leftrightarrow a=c\)
Chứng minh tương tự ta có: \(\hept{\begin{cases}b=c\\a=b\end{cases}}\)
\(\Rightarrow a=b=c=2\)
Vậy \(A_{min}=3\Leftrightarrow a=b=c=2\)
Câu 1:
x=79 nên x+1=80
\(A=1969-\left(x^{1969}-80x^{1968}+...-80x^2+80x\right)\)
\(=1969-\left[x^{1969}-x^{1968}\left(x+1\right)+...-x^2\left(x+1\right)+x\left(x+1\right)\right]\)
\(=1969-\left[x^{1969}-x^{1969}-x^{1968}+x^{1968}-...-x^3-x^2+x^2+x\right]\)
=1969-79=1890
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: \(a=b=c=\frac{1}{\sqrt[3]{3}}\)
\(VT=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\)
\(VT\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c\)
\(\Rightarrow\) Tam giác là tam giác đều
Lời giải:
Vì tam giác có chu vi bằng $1$ nên $a+b+c=1$
\(\Rightarrow 1-a, 1-b, 1-c>0\)
Thay vào biểu thức đã cho:
\(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}=\frac{3}{2}\)
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\right)[a(1-a)+b(1-b)+c(1-c)]\geq (a+b+c)^2\)
\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{(a+b+c)^2}{(a+b+c)-(a^2+b^2+c^2)}\)
\(\Leftrightarrow \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\)
Áp dụng BĐT Bunhiacopxky: \((a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2=1\)
\(\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}\)
\(\Rightarrow 1-(a^2+b^2+c^2)\leq \frac{2}{3}\)
Suy ra \( \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq \frac{1}{1-(a^2+b^2+c^2)}\geq \frac{1}{\frac{2}{3}}=\frac{3}{2}\)
Dấu bằng xảy ra khi \(\frac{a}{1}=\frac{b}{1}=\frac{c}{1}\Leftrightarrow a=b=c\) hay tam giác $ABC$ đều.
Cách khác:v
Giải: \(gt:\left\{{}\begin{matrix}a;b;c>0\\a+b+c=1\end{matrix}\right.\)
\(pt\Leftrightarrow\dfrac{a}{\left(a+b+c\right)-a}+\dfrac{b}{\left(a+b+c\right)-b}+\dfrac{c}{\left(a+b+c\right)-c}=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{3}{2}\)
\(Nesbit:\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Dấu "=" \(a=b=c\Leftrightarrow\Delta ABC\) đều