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Đề bài thiếu, chắc chắn phải có thêm 1 dữ kiện khác
Ví dụ, bạn cho \(a=b=c=1000\) sẽ thấy BĐT sai
\(A\le\frac{1}{27}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^3\)
Mặt khác:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{3\left[4\left(a+b+c\right)+3\right]}=3\sqrt{5}\)
\(\Rightarrow A\le\frac{1}{27}\left(3\sqrt{5}\right)^3=5\sqrt{5}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{\left(1+1+1\right)\left(4a+1+4b+1+4c+1\right)}\) \(=\sqrt{3.\left(4.3+3\right)}=\sqrt{3.15}=3\sqrt{5}\)
\(\text{Dấu ''='' xảy ra }\Leftrightarrow a=b=c=1\)
1)
\(2a+\frac{4}{a}+\frac{16}{a+2}=\left(a+\frac{4}{a}\right)+\left[\left(a+2\right)+\frac{16}{a+2}\right]-2\ge4+8-2=10\)
Dấu "=" xảy ra khi a=2
2)
\(\hept{\begin{cases}\sqrt{a\left(1-4a\right)}=\frac{1}{2}\sqrt{4a\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4a+1-4a}{2}=\frac{1}{4}\\\sqrt{b\left(1-4b\right)}=\frac{1}{2}\sqrt{4\left(1-4a\right)}\le\frac{1}{2}\cdot\frac{4b+1-4b}{2}=\frac{1}{4}\\\sqrt{c\left(1-4c\right)}=\frac{1}{2}\sqrt{4c\left(1-4c\right)}\le\frac{1}{2}\cdot\frac{4c+1-4c}{2}=\frac{1}{4}\end{cases}}\)
\(\Rightarrow\sqrt{a\left(1-4a\right)}+\sqrt{b\left(1-4b\right)}+\sqrt{c\left(1-4c\right)}\le\frac{3}{4}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{8}\)
\(P=\sum\frac{a}{\sqrt{\left(2a\right)^2+\left(b+c\right)^2}}\le\sqrt{2}\sum\frac{a}{2a+b+c}=\sqrt{2}\sum a\left(\frac{1}{a+b+a+c}\right)\le\frac{\sqrt{2}}{4}\sum\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3\sqrt{2}}{4}\)
Dấu "=" xảy ra khi \(a=b=c\)
Bài 1:
Đặt \(\begin{array}{l} \sqrt c = \alpha \\ \sqrt {b - c} = \beta \\ \end{array}\) và \(\begin{array}{l} \sqrt {a - c} = x \\ \sqrt c = y \\ \end{array}\)
Áp dụng bất đẳng thức Bunhiacopxki
\(x\alpha + y\beta \le \left| {x\alpha + y\beta } \right| \le \sqrt {{x^2} + {y^2}} .\sqrt {{\alpha ^2} + {\beta ^2}}\)
\(\begin{array}{l} \Leftrightarrow \sqrt c .\sqrt {a - c} + \sqrt {b - c} .\sqrt c \le \sqrt {{{\left( {\sqrt c } \right)}^2} + {{\left( {\sqrt {b - c} } \right)}^2}} .\sqrt {{{\left( {\sqrt c } \right)}^2} + {{\left( {\sqrt {a - c} } \right)}^2}} \\ \Leftrightarrow \sqrt {c(a - c)} + \sqrt {c(b - c)} \le \sqrt {c + (a - c)} .\sqrt {c + (b - c)} \\ \Leftrightarrow \sqrt {c(a - c)} + \sqrt {c(b - c)} \le \sqrt b \sqrt a = \sqrt {ab} . \\ \end{array}\)
P/s: Mình gõ latex kém quá khó hiểu chỗ não thì cứ hỏi :)))
Bài 2:
\(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)
\(\Leftrightarrow\sqrt{a}\sqrt{ab-a}+\sqrt{b}\sqrt{ab-b}\)\(\le\sqrt{\left(a+b\right)\left(2ab-a-b\right)}\)
\(\le\frac{a+b-a-b+2ab}{2}=ab\)
a)\(x -1 >5 ⇔ x > 1 ⇒ x^4 > x^3 > x^2 > x > 1 \)
\(⇒ 5x^4 > x^4 + x^3 + x^2 + x + 1 > 5 \)
\(⇒ 5x^4 (x-1) > (x-1)( x^4 + x^3 + x^2 + x + 1) = x^5 -1 > 5 (x-1) \)
b)\(x^5 + y^5 – x^4y – xy^4 = (x + y)(x^4 – x^3y + x^2y^2 – xy^3 + y^4) – xy(x^3 + y^3) \)
\(= (x + y) [( x^4 – x^3y+ x^2y^2 – xy^3 + y^4) – xy(x^2 – xy + y^2)] \)
\(= (x + y) [(x^4+2x^2y^2+y^4) - 2xy(x^2+y^2)] \)
\(= (x + y) (x - y)^2(x^2 + y^2) ≥ 0 \)
c)\(\sqrt {4a + 1} + \sqrt {4b + 1} + \sqrt {4c + 1} )^2\)
\(= 4(a + b + c) + 3 + 2\sqrt {4a + 1} \sqrt {4b + 1} + 2\sqrt {4a + 1} \sqrt {4c + 1} + 2\sqrt {4b + 1} \sqrt {4c + 1} \)
\( \le 4(a + b + c) + 3 + (4a + 1) + (4b + 1) + (4a + 1) + (4c + 1) + (4b + 1) + (4c + 1) \)
\(\le 12(a + b + c) + 9 \le 21 \le 25\)
we have that: \(\sqrt{a^4+b^2+c^2+1}=\sqrt{a^4-a^2+2}\)
and \(\dfrac{-a^2+11}{8}\le\sqrt{a^4-a^2+2}\le\sqrt{2}\) \(\left(a\in\left(0;1\right)\right)\)
Đặt:
\(A=\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\)
Áp dụng bất đẳng thức bunhiacopxki ta có:
\(A^2=\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)^2\le\left(1^2+1^2+1^2\right)\left(4a+1+4b+1+4c+1\right)=21\)
Hay \(A\le\sqrt{21}\left(đpcm\right)\)
Rảnh quá ủng hộ cách khác nè =))
Áp dụng Cô-si có:
\(\sqrt{4a+1}\cdot\sqrt{\dfrac{7}{3}}\le\dfrac{4a+1+\dfrac{7}{3}}{2}=2a+\dfrac{5}{3}\)
Tương tự vs 2 bđt còn lại: \(\left\{{}\begin{matrix}\sqrt{4b+1}\cdot\sqrt{\dfrac{7}{3}}\le2b+\dfrac{5}{3}\\\sqrt{4c+1}\cdot\sqrt{\dfrac{7}{3}}\le2c+\dfrac{5}{3}\end{matrix}\right.\)
Cộng 2 vế của 3 bđt trên có:
\(\sqrt{\dfrac{7}{3}}\left(\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\right)\le2\left(a+b+c\right)+5=7\)
\(\Leftrightarrow\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\le\sqrt{21}\)
Hoàn tất chứng minh