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Nice proof, nhưng đã quy đồng là phải thế này :v
\(BDT\Leftrightarrow\left(2a-\sqrt{a^2+3}\right)+\left(2b-\sqrt{b^2+3}\right)+\left(2c-\sqrt{c^2+3}\right)\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}\ge0\)
\(\Leftrightarrow\dfrac{a^2-1}{2a+\sqrt{a^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{a}-a\right)+\dfrac{b^2-1}{2b+\sqrt{b^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{b}-b\right)+\dfrac{c^2-1}{2c+\sqrt{c^2+3}}+\dfrac{1}{4}\left(\dfrac{1}{c}-c\right)\ge0\)
\(\Leftrightarrow\left(a^2-1\right)\left(\dfrac{1}{2a+\sqrt{a^2+3}}-\dfrac{1}{4a}\right)+\left(b^2-1\right)\left(\dfrac{1}{2b+\sqrt{b^2+3}}-\dfrac{1}{4b}\right)+\left(c^2-1\right)\left(\dfrac{1}{2c+\sqrt{a^2+3}}-\dfrac{1}{4c}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)\left(2a-\sqrt{a^2+3}\right)}{a\left(2a+\sqrt{a^2+3}\right)}+\dfrac{\left(b^2-1\right)\left(2b-\sqrt{b^2+3}\right)}{b\left(2b+\sqrt{b^2+3}\right)}+\dfrac{\left(c^2-1\right)\left(2c-\sqrt{c^2+3}\right)}{c\left(2c+\sqrt{c^2+3}\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a^2-1\right)^2}{a\left(2a+\sqrt{a^2+3}\right)^2}+\dfrac{\left(b^2-1\right)^2}{b\left(2b+\sqrt{b^2+3}\right)^2}+\dfrac{\left(c^2-1\right)^2}{c\left(2c+\sqrt{c^2+3}\right)^2}\ge0\) (luôn đúng)
Khi \(f\left(t\right)=\sqrt{1+t}\) là hàm lõm trên \([-1, +\infty)\) ta có:
\(f(t)\le f(3)+f'(3)(t-3)\forall t\ge -1\)
Tức là \(f\left(t\right)\le2+\dfrac{1}{4}\left(t-3\right)=\dfrac{5}{4}+\dfrac{1}{4}t\forall t\ge-1\)
Áp dụng BĐT này ta có:
\(\sqrt{a^2+3}=a\sqrt{1+\dfrac{3}{a^2}}\le a\left(\dfrac{5}{4}+\dfrac{1}{4}\cdot\dfrac{3}{a^2}\right)=\dfrac{5}{4}a+\dfrac{3}{4}\cdot\dfrac{1}{a}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\sqrt{b^2+3}\le\dfrac{5}{4}b+\dfrac{3}{4}\cdot\dfrac{1}{b};\sqrt{c^2+3}\le\dfrac{5}{4}c+\dfrac{3}{4}\cdot\dfrac{1}{c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VP\le\dfrac{5}{4}\left(a+b+c\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2\left(a+b+c\right)=VT\)
Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)
\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)
CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)
Cộng vế theo vế:
\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)
Dấu \("="\Leftrightarrow a=b=c=2\)
ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có
\(\Sigma\dfrac{1}{\left(a+b\right)c}\le\Sigma\dfrac{1}{2\sqrt{c}}=\dfrac{\Sigma\sqrt{ab}}{2}\le\dfrac{\Sigma a}{2}\)(đpcm)
Đây là BĐT Iran 96 khá nổi tiếng. Bạn hoàn toàn có thể search trên google lời giải.
Đây nhá:)Sửa đề:
Chứng minh rằng \(\Sigma S_a\left(b-c\right)^2\ge S\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Nếu \(s_a+S_b\ge0;S_b+S_c\ge0;2\sqrt{\left(S_a+S_b\right)\left(S_b+S_c\right)}+2S_b-S\left(c-a\right)\ge0\)
Xét TH \(a\ge b\ge c\) thì bđt đề bài hiển nhiên đúng nên ta chỉ xét:
\(a\le b\le c\) khi đó \(\left(a-b\right)\left(b-c\right)\ge0\) (1)
Ta có: \(S_a\left(b-c\right)^2+S_b\left(c-a\right)^2+S_c\left(a-b\right)^2\)
\(=\left(S_a+S_b\right)\left(b-c\right)^2+\left(S_c+S_b\right)\left(a-b\right)^2+2S_b\left(a-b\right)\left(b-c\right)\)
\(\ge2\sqrt{\left(S_a+S_b\right)\left(S_b+S_c\right)}\left(a-b\right)\left(b-c\right)+2S_b\left(b-c\right)\left(a-b\right)\)
(CÔ si)
Như vậy, BĐT đề bài sẽ được chứng minh nếu:
\(2\sqrt{\left(S_a+S_b\right)\left(S_b+S_c\right)}\left(a-b\right)\left(b-c\right)+2S_b\left(b-c\right)\left(a-b\right)\ge S\left(a-b\right)\left(b-c\right)\left(c-a\right)\)\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(2\sqrt{\left(S_a+S_b\right)\left(S_b+S_c\right)}+2S_b-S\left(c-a\right)\right)\ge0\)
Và điều này luôn đúng theo (1) và giả thiết đề bài.
\(S_a\left(b-c\right)^2\) là gì vậy, cái này em chưa học. Giải thích đi để em xem thế nào...
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 $a=b=c$
Từ \(a^2+b^2+c^2=3\Rightarrow a+b+c\le3\)
Ta có: \(\sqrt{\dfrac{9}{\left(a+b\right)^2}+c^2}+\sqrt{\dfrac{9}{\left(b+c\right)^2}+a^2}+\sqrt{\dfrac{9}{\left(c+a\right)^2}+b^2}\)
\(\ge\sqrt{9\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)^2+\left(a+b+c\right)^2}\)
\(\ge\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\)
Cần chứng minh \(\sqrt{9\cdot\left(\dfrac{9}{2\left(a+b+c\right)}\right)^2+\left(a+b+c\right)^2}\ge\dfrac{3\sqrt{13}}{2}\)
\(\Leftrightarrow9\left(\dfrac{9}{2t}\right)^2+t^2\ge\dfrac{117}{4}\left(t=a+b+c\le3\right)\)
\(\Leftrightarrow\dfrac{\left(t-3\right)\left(2t-9\right)\left(t+3\right)\left(2t+9\right)}{4t^2}\ge0\)*Đúng*
B1:a)ĐK: \(x\ne 0;4;9\)
b)\(P=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\sqrt{x}-1+1}{\sqrt{x}+1}\right)\)
\(=\dfrac{x-9-x+4+x^{\dfrac{1}{2}}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{x^{\dfrac{1}{2}}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)
\(=\dfrac{1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1}{x^{\dfrac{1}{2}}}\)\(=\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}\)
c)Vì \(x^{\dfrac{1}{2}}+1>0\forall x\) nên
\(P< 0< =>x-2x^{\dfrac{1}{2}}< 0\)
\(\Leftrightarrow x^{\dfrac{1}{2}}\left(x^{\dfrac{1}{2}}-2\right)< 0\)
\(\Leftrightarrow0< x< 4\)
Vậy 0<x<4 thì P<0
d)tA CÓ: \(\dfrac{1}{P}=\dfrac{x-2x^{\dfrac{1}{2}}}{x^{\dfrac{1}{2}}+1}=\dfrac{x-2x^{\dfrac{1}{2}}+1-1}{x^{\dfrac{1}{2}}+1}=\dfrac{\left(x^{\dfrac{1}{2}}-1\right)^2-1}{x^{\dfrac{1}{2}}+1}\ge-1\)
"=" khi x=1
B2:
a)\(A=x^2-2xy+y^2+4x-4y-5\)
\(=\left(x-y\right)^2+4\left(x-y\right)-5\)
\(=\left(x-y\right)^2-1+4\left(x-y\right)-4\)
\(=\left(x-y+1\right)\left(x-y-1\right)+4\left(x-y-1\right)\)
\(=\left(x-y+5\right)\left(x-y-1\right)\)
b)\(P=x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+2x^3+x^2\right)+2\left(x^2+x\right)+1\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1\)
\(=\left(x^2+x+1\right)^2\ge0\forall x\)
Vậy MinP=0
c)\(Q=x^6+2x^5+2x^4+2x^3+2x^2+2x+1\)
\(=\left(x^2+x-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=\left(1-1\right)\left(x^4+x^3+2x^2+x+3\right)+4\)
\(=0\left(x^4+x^3+2x^2+x+3\right)+4=4\)
Vậy x^2+x=1 thì Q=4
B3:a)\(2xy+x+y=83\)
\(\Leftrightarrow x\left(2y+1\right)+\dfrac{1}{2}\left(2y+1\right)=\dfrac{167}{2}\)
\(\Leftrightarrow2x\left(2y+1\right)+1\left(2y+1\right)=167\)
\(\Leftrightarrow\left(2x+1\right)\left(2y+1\right)=167\)
Mà \(Ư\left(167\right)=\left\{\pm1;\pm167\right\}\)
\(\Leftrightarrow\left(x;y\right)=\left(-84;-1\right);\left(-1;-84\right);\left(0;83\right);\left(83;0\right)\)
Vậy...
b)\(y^2+2xy-3x-2=0\)
\(\Leftrightarrow x^2+y^2+2xy-x^2-3x-2=0\)
\(\Leftrightarrow\left(x+y\right)^2=x^2+3x+2\)
\(\Leftrightarrow\left(x+y\right)^2=\left(x+1\right)\left(x+2\right)\)
Vì \(x;y\in Z\) nên VT là số chính phương VP là tích 2 số nguyên liên tiếp
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=2\end{matrix}\right.\)
Vậy...
B5:\(B=\dfrac{x^2+x+1}{x^2-x+1}\)
\(\Leftrightarrow x^2\left(B-1\right)+x\left(-B-1\right)+\left(B-1\right)=0\)
\(\Delta=\left(-B-1\right)^2-4\left(B-1\right)\left(B-1\right)\)
\(=-\left(B-3\right)\left(3B-1\right)\)
pt có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(B-3\right)\left(3B-1\right)\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}B-3\le0\\3B-1\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}B\le3\\B\ge\dfrac{1}{3}\end{matrix}\right.\)
Min B=1/3 khi x=-1; Max B=3 khi x=1
Đề bắt quy nạp khó quá, giá đề mở thì xài Ber's ineq cho lẹ .-.
*) Với \(n=1;2\) BĐT đúng
*)Giả sử BĐT đúng với \(n=k\) tức chứng minh BĐT đúng với \(n=k+1\) hay \(\dfrac{a^{k+1}+b^{k+1}}{2}\ge\left(\dfrac{a+b}{2}\right)^{k+1}\)
Ta có: \(VT-VP=\dfrac{a^{k+1}+b^{k+1}}{2}-\left(\dfrac{a+b}{2}\right)^{k+1}\)
\(=\dfrac{a^{k+1}+b^{k+1}}{2}-\left(\dfrac{a+b}{2}\right)^k\left(\dfrac{a+b}{2}\right)\)
\(\ge\dfrac{a^{k+1}+b^{k+1}}{2}-\dfrac{a^k+b^k}{2}\cdot\dfrac{a+b}{2}\)
\(=\dfrac{\left(a-b\right)\left(a^k-b^k\right)}{4}=\dfrac{\left(a-b\right)\left(a^{k-1}-a^{k-2}b+...+b^{k-1}\right)}{4}\ge0\)
Khi \(a=b\)