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Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng
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 : \(\Sigma\dfrac{1}{\sqrt{2\left(a^2+b^2\right)}+4}\le2.\dfrac{1}{4}\left(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\right)=\dfrac{1}{2}.1=\dfrac{1}{2}\)
" = " \(\Leftrightarrow a=b=c=1\)
\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)
\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)
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 $a=b=c$
`sqrta+sqrtb+sqrtc=2`
`<=>(sqrta+sqrtb+sqrtc)^2=4`
`<=>a+b+c+2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4`
`<=>2sqrt{ab}+2sqrt{bc}+2sqrt{ca}=4-(a+b+c)=4-2-2`
`<=>sqrt{ab}+sqrt{bc}+sqrt{ca}=1`
`=>a+1=a+sqrt{ab}+sqrt{bc}+sqrt{ca}=sqrta(sqrta+sqrtb)+sqrtc(sqrta+sqrtb)=(sqrta+sqrtb)(sqrta+sqrtc)`
Tương tự:`b+1=(sqrtb+sqrta)(sqrtb+sqrtc)`
`c+1=(sqrtc+sqrta)(sqrtc+sqrtb)`
`=>VT=sqrta/((sqrta+sqrtb)(sqrta+sqrtc))+sqrtb/((sqrtb+sqrta)(sqrtb+sqrtc))+sqrtc/((sqrtc+sqrta)(sqrtc+sqrtb))`
`=>VT=(sqrta(sqrtb+sqrtc)+sqrtb(sqrtc+sqrta)+sqrtc(sqrta+sqrtb))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=(sqrt{ab}+sqrt{ac}+sqrt{bc}+sqrt{ab}+sqrt{ac}+sqrt{bc})/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=(2(sqrt{ab}+sqrt{bc}+sqrt{ca}))/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=2/((sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta))`
`=2/\sqrt{[(sqrta+sqrtb)(sqrtb+sqrtc)(sqrtc+sqrta)]^2}`
`=2/\sqrt{(sqrta+sqrtb)(sqrta+sqrtc)(sqrtb+sqrta)(sqrtb+sqrtc)(sqrtc+sqrta)(sqrtc+sqrtb)}`
`=2/\sqrt{(1+a)(1+b)(1+c)}=>đpcm`
a ơi giả thiết là a+b+c=\(\sqrt{a}+\sqrt{b}+\sqrt{c}\)=2 nhé a
Ta có \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\)
\(=\sqrt{2a\left(a+b+c\right)+\dfrac{b^2-2bc+c^2}{2}}\)
\(=\sqrt{\dfrac{4a^2+b^2+c^2+4ab+4ac-2bc}{2}}\)
\(=\sqrt{\dfrac{\left(2a+b+c\right)^2-4bc}{2}}\)
\(\le\sqrt{\dfrac{\left(2a+b+c\right)^2}{2}}\)
\(=\dfrac{2a+b+c}{\sqrt{2}}\).
Vậy \(\sqrt{2022a+\dfrac{\left(b-c\right)^2}{2}}\le\dfrac{2a+b+c}{\sqrt{2}}\). Lập 2 BĐT tương tự rồi cộng vế, ta được \(VT\le\dfrac{2a+b+c+2b+c+a+2c+a+b}{\sqrt{2}}\)
\(=\dfrac{4\left(a+b+c\right)}{\sqrt{2}}\) \(=\dfrac{4.1011}{\sqrt{2}}\) \(=2022\sqrt{2}\)
ĐTXR \(\Leftrightarrow\) \(\left\{{}\begin{matrix}ab=0\\bc=0\\ca=0\\a+b+c=1011\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(1011;0;0\right)\) hoặc các hoán vị. Vậy ta có đpcm.
Ta có:
\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=9\\ \Leftrightarrow a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}=9\\ \Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=2\)
\(\Rightarrow\dfrac{\sqrt{a}}{a+2}+\dfrac{\sqrt{b}}{b+2}+\dfrac{\sqrt{c}}{c+2}=\dfrac{\sqrt{a}}{a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{b}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}+\dfrac{\sqrt{c}}{c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}\\ =\dfrac{\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{c}\right)}+\dfrac{\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}+\dfrac{\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{\sqrt{a}\left(\sqrt{b}+\sqrt{c}\right)+\sqrt{b}\left(\sqrt{a}+\sqrt{c}\right)+\sqrt{c}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}\\ =\dfrac{4}{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{b}+\sqrt{c}\right)^2\left(\sqrt{a}+\sqrt{c}\right)^2}}\)\(=\dfrac{4}{\sqrt{\left(a+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\left(c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)}}\\ =\dfrac{4}{\sqrt{\left(a+2\right)\left(b+2\right)\left(c+2\right)}}\)
Từng sau nếu tag bạn tag tên dưới câu trả lời nhé, tag thế này không nhận được thông báo đâu .
Bài này tốn sức quá, đau đầu
Lời giải:
Sử dụng \(\sum\) biểu hiện tổng các hoán vị nhé.
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{a^2}{a\sqrt{(b+2)(c+2)}}+\frac{b^2}{b\sqrt{(c+2)(a+2)}}+\frac{c^2}{c\sqrt{(a+2)(b+2)}}\geq \frac{(a+b+c)^2}{\sum a\sqrt{(b+2)(c+2)}}\)
Tiếp tục Cauchy-Schwarz:
\((\sum a\sqrt{(b+2)(c+2)})^2\leq (ab+2a+bc+2b+ac+2c)(ac+2a+ba+2b+bc+2c)\)
\(\Leftrightarrow \sum a\sqrt{(b+2)(c+2)}\leq (ab+bc+ac+2a+2b+2c)\)
\(\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{ab+bc+ac+2(a+b+c)}\)
Ta sẽ đi chứng minh \(\frac{(a+b+c)^2}{ab+bc+ac+2(a+b+c)}\geq 1\Leftrightarrow (a+b+c)^2\geq ab+bc+ac+2(a+b+c)\)
\(\Leftrightarrow a^2+b^2+c^2+ab+bc+ac\geq 2(a+b+c)\)
\(\Leftrightarrow (a^2+b^2+c^2)+(a+b+c)^2\geq 4(a+b+c)\)
\(\Leftrightarrow 4-abc+(a+b+c)^2\geq 4(a+b+c)\Leftrightarrow (a+b+c-2)^2\geq abc\)
\(\Leftrightarrow a+b+c\geq \sqrt{abc}+2\)
Do \(a^2+b^2+c^2+abc=4\Rightarrow \)
tồn tại $x,y,z>0$ sao cho:\((a,b,c)=\left ( 2\sqrt{\frac{xy}{(z+x)(z+y)}};2\sqrt{\frac{yz}{(x+y)(x+z)}};2\sqrt{\frac{xz}{(y+x)(y+z)}} \right )\)
Khi đó , thực hiện vài bước rút gọn, BĐT cần chứng minh chuyển về:
\(\sum \sqrt{xy(x+y)}\geq \sqrt{2xyz}+\sqrt{(x+y)(y+z)(x+z)}\)
Bình phương hai vế:
\(\Leftrightarrow \sum xy(x+y)+2\sqrt{xy^2z(x+y)(y+z)}\geq 2xyz+\prod (x+y)+2\sqrt{2xyz(x+y)(y+z)(x+z)}\)
\(\Leftrightarrow \sum\sqrt{xy^2z(x+y)(y+z)}\geq 2xyz+\sqrt{2xyz(x+y)(y+z)(x+z)}\)
\(\Leftrightarrow \sum \sqrt{y(y+x)(y+z)}\geq 2\sqrt{xyz}+\sqrt{2(x+y)(y+z)(x+z)}\) \((\star)\)
Đặt biểu thức vế trái là $A$
\(A^2=\sum y(y+x)(y+z)+2\sum\sqrt{[y(y+x)(y+z)][x(x+y)(x+z)]}\)
\(A^2=\sum x^3+\sum xy(x+y)+3xyz+2\sum \sqrt{[(x^2(x+y+z)+xyz][y^2(x+y+z)+xyz]}\)
Áp dụng BĐT C-S : \([x^2(x+y+z)+xyz][y^2(x+y+z)+xyz]\geq [xy(x+y+z)+xyz]^2\)
\(\Rightarrow A^2\geq \sum x^3+\sum xy(x+y)+3xyz+2\sum [xy(x+y+z)+xyz]\)
\(\Leftrightarrow A^2\geq \sum x^3+3\sum xy(x+y)+15xyz\)
Theo BĐT Schur: \(\sum x^3+3xyz\geq \sum xy(x+y)\)
\(\Rightarrow A^2\geq 4\sum xy(x+y)+12xyz=4[\sum xy(x+y)+3xyz]=4(x+y+z)(xy+yz+xz)\)
\(\Leftrightarrow A\geq 2\sqrt{(x+y+z)(xy+yz+xz)}\)
Ta cần chứng minh \(2\sqrt{(x+y+z)(xy+yz+xz)}\geq 2\sqrt{xyz}+\sqrt{2(x+y)(y+z)(x+z)}\) (1)
Đặt \(\sqrt{(x+y+z)(xy+yz+xz)}=t\), bằng AM-GM dễ thấy \(t^2\geq 9xyz\)
\((1)\Leftrightarrow 2t\geq 2\sqrt{xyz}+\sqrt{2(t^2-xyz)}\)
\(\Leftrightarrow 4t^2\geq 4xyz+2(t^2-xyz)+4\sqrt{2xyz(t^2-xyz)}\)
\(\Leftrightarrow t^2\geq xyz+2\sqrt{2xyz(t^2-xyz)}\) (2)
Áp dụng AM-GM: \(2\sqrt{xyz(t^2-xyz)}=\sqrt{8xyz(t^2-xyz)}\leq \frac{8xyz+t^2-xyz}{2}=\frac{7}{2}xyz+\frac{t^2}{2}\)
Và \(xyz\leq \frac{t^2}{9}\)
\(\Rightarrow xyz+2\sqrt{2xyz(t^2-xyz)}\leq t^2\)
Do đó (2) đúng kéo theo (1) đúng kéo theo (*) đúng nên ta có đpcm.
Dấu bằng xảy ra khi $a=b=c=1$
Thánh giỏi quá. Em xin bái phục!