Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(a+b+c+\sqrt{abc}=4\)
\(\Rightarrow4a+4b+4c+4\sqrt{abc}=16\)
\(\Rightarrow4a+4\sqrt{abc}=16-4b-4c\)
\(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(16-4b-4c+bc\right)}=\sqrt{a\left(4a+4\sqrt{abc}+bc\right)}\)
\(=\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}=\left|2a+\sqrt{abc}\right|=2a+\sqrt{abc}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{b\left(4-a\right)\left(4-c\right)}=2b+\sqrt{abc}\\\sqrt{c\left(4-a\right)\left(4-b\right)}=2c+\sqrt{abc}\end{matrix}\right.\)
\(\Rightarrow A=\sqrt{a\left(4-b\right)\left(4-c\right)}+\sqrt{b\left(4-c\right)\left(4-a\right)}+\sqrt{c\left(4-a\right)\left(4-b\right)}-\sqrt{abc}=2a+2b+2c+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c+\sqrt{abc}\right)=8\)
Ta có \(\sqrt{a\left(4-b\right)\left(4-c\right)}=\sqrt{a\left(a+c+\sqrt{abc}\right)\left(4-c\right)}\)
\(=\sqrt{\left(a^2+ac+a\sqrt{abc}\right)\left(4-c\right)}\\ =\sqrt{4a^2+ac\left(4-\sqrt{abc}-a-c\right)+4a\sqrt{abc}}\\ =\sqrt{4a^2+4a\sqrt{abc}+abc}=\sqrt{\left(2a+\sqrt{abc}\right)^2}\\ =2a+\sqrt{abc}\left(a,b,c>0\right)\)
Cmtt \(\sqrt{b\left(4-c\right)\left(4-a\right)}=2b+\sqrt{abc};\sqrt{c\left(4-b\right)\left(4-a\right)}=2c+\sqrt{abc}\)
\(\Rightarrow A=2\left(a+b+c\right)+3\sqrt{abc}-\sqrt{abc}=2\left(a+b+c\right)+2\sqrt{abc}\\ A=2\left(a+b+c+\sqrt{abc}\right)=2\cdot4=8\)
Ta có \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Để tổng trên chia hết cho 81 thì \(\left(a-b\right)\left(b-c\right)\left(c-a\right)⋮27\)
Mà \(a+b+c=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Bài toán trở thành: Cho \(x+y+z=\left(x-y\right)\left(y-z\right)\left(z-x\right)\). CMR: \(x+y+z⋮27\) - Hoc24
Lời giải:
Áp dụng BĐT AM-GM:
$\frac{a^3}{(a+b)(a+c)}+\frac{a+b}{8}+\frac{a+c}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3}{4}a$
$\frac{b^3}{(b+c)(b+a)}+\frac{b+c}{8}+\frac{b+a}{8}\geq \frac{3}{4}b$
$\frac{c^3}{(c+a)(c+b)}+\frac{c+a}{8}+\frac{c+b}{8}\geq \frac{3}{4}c$
Cộng 3 BĐT trên và thu gọn:
$\Rightarrow \frac{a^3}{(a+b)(a+c)}+\frac{b^3}{(b+a)(b+c)}+\frac{c^3}{(c+a)(c+b)}\geq \frac{1}{4}(a+b+c)=\frac{1}{4}.3=\frac{3}{4}$
Vậy ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
Bạn tham khảo tại đây:
Câu hỏi của Trần Hữu Ngọc Minh - Toán lớp 9 - Học toán với OnlineMath
Áp dụng BĐT Cosi ta được:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)64}}=\frac{3a}{4}̸\)
Tương tự \(\hept{\begin{cases}\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{1+a}{8}+\frac{1+c}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3c}{4}\end{cases}}\)
Cộng theo từng vế BĐT trên ta có:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{a+b+c}{2}\)
Vì \(a+b+c\ge3\sqrt[3]{abc}=3\)do đó:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{3}{4}\left(đpcm\right)\)
Đẳng thức xảy ra <=> a=b=c
Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b+2}{36}+\frac{c+3}{48}\geq 3\sqrt[3]{\frac{a^3}{36.48}}=\frac{a}{4}\)
Tương tự:\(\frac{b^3}{(c+2)(a+3)}+\frac{c+2}{36}+\frac{a+3}{48}\geq \frac{b}{4}\)
\(\frac{c^3}{(a+2)(b+3)}+\frac{a+2}{36}+\frac{b+3}{48}\geq \frac{c}{4}\)
Cộng theo vế các BĐT trên và rút gọn ta có:
\(\frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\)
Mà cũng theo AM-GM:
\(a+b+c\geq 3\sqrt[3]{abc}=3\)
\(\Rightarrow \frac{a^3}{(b+2)(c+3)}+\frac{b^3}{(c+2)(a+3)}+\frac{c^3}{(a+2)(b+3)}\geq \frac{29}{144}(a+b+c)-\frac{17}{48}\geq \frac{29}{144}.3-\frac{17}{48}=\frac{1}{4}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
bđt trái dấu rồi nha!
\(P=\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{c^3}{\left(a+1\right)\left(b+1\right)}\ge\frac{3}{4}\)
+ Áp dụng bđt Cauchy ta có :
\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{8}+\frac{c+1}{8}\ge3\sqrt[3]{\frac{a^3}{\left(b+1\right)\left(c+1\right)}\cdot\frac{b+1}{8}\cdot\frac{c+1}{8}}=\frac{3}{4}a\). Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}2a=b+1\\b=c\end{matrix}\right.\)
+ Tương tự ta c/m đc : \(\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{a+1}{8}+\frac{c+1}{8}\ge\frac{3}{4}b\). Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}2b=a+1\\a=c\end{matrix}\right.\)
\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{b+1}{8}\ge\frac{3}{4}c\). Dấu "=" \(\Leftrightarrow2c=a+1=b+1\)
Do đó : \(P\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(a+b+c\right)-\frac{3}{4}=\frac{1}{2}\left(a+b+c\right)-\frac{3}{4}\) \(\ge\frac{1}{2}\cdot3\sqrt[3]{abc}-\frac{3}{4}=\frac{3}{4}\)
Dấu "=" \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b+1}{8}+\dfrac{c+1}{8}\)
\(\ge3\sqrt[3]{\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}\cdot\dfrac{b+1}{8}\cdot\dfrac{c+1}{8}}=\dfrac{3a}{4}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{\left(1+c\right)\left(1+a\right)}+\dfrac{c+1}{8}+\dfrac{a+1}{8}\ge\dfrac{3b}{4};\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}+\dfrac{a+1}{8}+\dfrac{b+1}{8}\ge\dfrac{3c}{4}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{2\left(a+b+c+3\right)}{8}\ge\dfrac{3\left(a+b+c\right)}{4}\)
\(\Leftrightarrow VT+\dfrac{2\left(3\sqrt[3]{abc}+3\right)}{8}\ge\dfrac{3\cdot3\sqrt[3]{abc}}{4}\Leftrightarrow VT\ge\dfrac{3}{4}=VP\)
Khi \(a=b=c=1\)
\(\left(a+b+c\right)^3=\left(a+b+c\right)\left[\left(a+b-c\right)^2+\left(b+c-a\right)^2+\left(c+a-b\right)^2-\left(a+b-c\right)\left(b+c-a\right)-\left(b+c-a\right)\left(c+a-b\right)-\left(a+b-c\right)\left(c+a-b\right)\right]\)\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-\left(a+b-c\right)^2-\left(b+c-a\right)^2-\left(c+a-b\right)^2+\left(a+b-c\right)\left(b+c-a\right)+\left(b+c-a\right)\left(c+a-b\right)+\left(a+b-c\right)\left(c+a-b\right)\right]=0\)\(\Leftrightarrow\left(a+b+c\right)\left[4ac+4bc-2\left(a^2-2ab+b^2\right)-2c^2-\left(a^2-2ac+c^2\right)+b^2-\left(a^2-2ab+b^2\right)+c^2-\left(b^2-2bc+c^2\right)+a^2\right]=0\)\(\Leftrightarrow\left(a+b+c\right)\left[a^2+b^2+c^2-2ab-2bc-2ac\right]=0\)
Đến đây bí thật rồi TT
mày bị thần kinh à