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Bài 1:
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\) với a,b,c > 0
Áp dụng BĐT Chauchy cho 2 số không âm, ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge c\sqrt{\dfrac{b}{a}.\dfrac{a}{b}}=2c\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}=a\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\ge a\sqrt{\dfrac{c}{b}.\dfrac{b}{c}}=2a\)
\(\dfrac{ab}{c}+\dfrac{bc}{a}=b\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge b\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2b\)
Cộng vế theo vế ta được:
\(2\left(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
5)
a)
Có 3x+y = 1
\(\Rightarrow x+x+x+y=1\)
Áp dụng bất đẳng thức bunhiacopxki ta có :
\(\left(x^2+x^2+x^2+y^2\right)\left(1^2+1^2+1^2+1^2\right)\ge\left(x+x+x+y\right)^2\)
\(\Rightarrow3x^2+y^{2^{ }}.4\ge\left(3x+y\right)^2\)
\(\Rightarrow3x^2+y^2\ge\dfrac{1}{4}\)
b)
Áp dụng bất đẳng thức AM - GM ta có :
\(\left[{}\begin{matrix}a^2+1^2\ge2a\\b^2+1^2\ge2b\\c^2+1^2\ge2c\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(a+1\right)^2\ge4a^{ }\\\left(b+1\right)^2\ge4b^{ }\\\left(c+1\right)^2\ge4c^{ }\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge4a^{ }.4b.4c^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64a^{ }bc^{ }\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64abc\)
\(\Rightarrow\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge64\)
\(\Rightarrow\left(a+1\right)^{ }\left(b+1\right)^{ }\left(c+1\right)^{ }\ge8\) \(\left(đpcm\right)\)
3)
Sửa đề \(A=\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
Đặt b + c - a = x , a+c-b = y , a+b-c= z
\(\Rightarrow\left[{}\begin{matrix}2a=y+z\\2b=x+z\\2c=x+y\end{matrix}\right.\)
Có :
\(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
\(\Rightarrow\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Rightarrow\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\)
Áp dụng bất đẳng thức \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\forall a,b>0\)
\(\Rightarrow\) \(\left(\dfrac{y}{x}+\dfrac{x}{y}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)\ge6\)
\(\Rightarrow\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6\)
\(\Rightarrow2\left(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\right)\ge6\)
\(\Rightarrow\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\) \(\left(đpcm\right)\)
Bài 2:
a, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}+\dfrac{3x+1}{x^2-1}\right).\dfrac{x^2-1}{2x+1}\)
\(P=\dfrac{\left(x-1\right)^2-x\left(x+1\right)+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(P=\dfrac{2}{2x+1}\)
b, ĐKXĐ: \(x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P=\dfrac{3}{x-1}\Leftrightarrow\dfrac{2}{2x+1}=\dfrac{3}{x-1}\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
\(\Leftrightarrow2x-2=6x+3\)\(\Leftrightarrow-4x=5\Leftrightarrow x=\dfrac{-5}{4}\)(TMĐK)
c, \(ĐKXĐ:x\ne\pm1;x\ne\dfrac{-1}{2}\)
Để \(P\in Z\Leftrightarrow\dfrac{2}{2x+1}\in Z\Leftrightarrow2x+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
+) Với \(2x+1=1\Leftrightarrow x=0\left(TMĐK\right)\)
+) Với \(2x+1=-1\Leftrightarrow x=-1\left(KTMĐK\right)\)
+) Với \(2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)
+) Với \(2x+1=-2\Leftrightarrow x=\dfrac{-3}{2}\left(TMĐK\right)\)
Vậy để \(P\in Z\Leftrightarrow x\in\left\{0;\dfrac{1}{2};\dfrac{-3}{2}\right\}\)
Lời giải:
Áp dụng BĐT AM-GM: \(a^2+b^2\geq 2ab\Rightarrow 2(a^2+b^2)\geq (a+b)^2\)
\(a+b=a^2+b^2\geq \frac{(a+b)^2}{2}\Rightarrow 2(a+b)\geq (a+b)^2\)
Do đó mà \(a+b\leq 2\)
Có:
\(S=\frac{a}{a+1}+\frac{b}{b+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}=2-\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\)
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a+1}+\frac{1}{b+1}\geq \frac{4}{a+b+2}\geq \frac{4}{2+2}=1\) do \(a+b\leq 2\)
Do đó: \(S\leq 2-1\Leftrightarrow S\leq 1\)
Vậy \(S_{\max}=1\Leftrightarrow a=b=1\)
\(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\)
Áp dụng BĐT Cauchy-Schwarz, ta có:
\(\dfrac{a^2}{a^2+a}+\dfrac{b^2}{b^2+b}\ge\dfrac{\left(a+b\right)^2}{a^2+b^2+a+b}=\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\)
Bởi vì:\(a^2+b^2=a+b\Rightarrow a^2+b^2+a+b=a^2+b^2+a^2+b^2=2\left(a^2+b^2\right)\)
Mặt khác, theo Bunyakovsky, ta có:
\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
Do đó: \(\dfrac{\left(a+b\right)^2}{2\left(a^2+b^2\right)}\le\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2}=1\)
Vậy: \(Max_S=1\Leftrightarrow a=b=1\)
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
3.
\(A=\dfrac{2x+1}{x^2+2}=\dfrac{x^2+2-x^2+2x-1}{x^2+2}=\dfrac{\left(x^2+2\right)-\left(x^2-2x+1\right)}{x^2+2}=1-\dfrac{\left(x-1\right)^2}{x^2+2}\)
Ta có: \(\dfrac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\in R\)
⇒ \(A=1-\dfrac{\left(x-1\right)^2}{x^2+2}\le1\)
Vậy: \(Max_A=1\Leftrightarrow x=1\)
* \(A=\dfrac{2x+1}{x^2+2}=\dfrac{2\left(2x+1\right)}{2\left(x^2+2\right)}=\dfrac{4x+2}{2\left(x^2+2\right)}=\dfrac{-x^2-2+x^2+4x+4}{2\left(x^2+2\right)}\)
\(=-\dfrac{1}{2}+\dfrac{x^2+4x+4}{x^2+2}=-\dfrac{1}{2}+\dfrac{\left(x+2\right)^2}{x^2+2}\ge-\dfrac{1}{2}\)
Vậy: \(Min_A=-\dfrac{1}{2}\Leftrightarrow x=-2\)
* \(B=\dfrac{4x+3}{x^2+1}\) ( 1 cách khác)
\(\Rightarrow B\left(x^2+1\right)=4x+3\)
\(\Rightarrow Bx^2-4x+B-3=0\) (1) \(\left(a=B;b=-4,c=B-3\right)\)
* Với B = 0, pt (1) có nghiệm x = \(-\dfrac{3}{4}\)
* Với B ≠ 0, pt (1) có nghiệm khi và chỉ khi:
\(\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(-4\right)^2-4.B.\left(B-3\right)\ge0\)
\(\Rightarrow16-4B^2+12B\ge0\)
\(\Rightarrow\left(B-4\right)\left(B+1\right)\ge0\)
\(\Rightarrow-1\le B\le4\)
Suy ra: \(Min_B=-1\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.\left(-1\right)}=-2\)
\(Max_B=4\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{4}{2.4}=\dfrac{1}{2}\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{ac}+\dfrac{2}{bc}=4\)
<=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) +\(2\left(\dfrac{c}{abc}+\dfrac{b}{abc}+\dfrac{a}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{a+b+c}{abc}\right)=4\)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{abc}{abc}\right)=4\) (vì a+b+c =abc)
<=> \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)
2a)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2a+b+c}=\dfrac{1}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\\dfrac{1}{a+2b+c}=\dfrac{1}{a+b+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\\\dfrac{1}{a+b+2c}=\dfrac{1}{a+c+b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+\dfrac{1}{4}\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)+\dfrac{1}{4}\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Rightarrow VT\le\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{4\left(a+b\right)}+\dfrac{1}{4\left(a+c\right)}+\dfrac{1}{4\left(b+c\right)}\)
\(\Rightarrow VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
Chứng minh rằng \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\\\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )
Vì \(\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Mà \(VT\le\dfrac{1}{2\left(a+b\right)}+\dfrac{1}{2\left(b+c\right)}+\dfrac{1}{2\left(c+a\right)}\)
\(\Rightarrow\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
2b)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+a^2\ge2\sqrt{a^2}=2a\\1+b^2\ge2\sqrt{b^2}=2b\\1+c^2\ge2\sqrt{c^2}=2c\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{1+a^2}\le\dfrac{a}{2a}=\dfrac{1}{2}\\\dfrac{b}{1+b^2}\le\dfrac{b}{2b}=\dfrac{1}{2}\\\dfrac{c}{1+c^2}\le\dfrac{c}{2c}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{1+a^2}+\dfrac{b}{1+b^2}+\dfrac{c}{1+c^2}\le\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Bài 1)
Nháp : nhìn nhanh ta thấy nên áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Giải
Vì x,y > 0 =) 2x + y > 0 , x + 2y > 0
Áp dụng BĐT cauchy dạng phân thức cho hai bộ số không âm \(\dfrac{1}{2x+y}\)và\(\dfrac{1}{x+2y}\)
\(\Rightarrow\dfrac{1}{x+2y}+\dfrac{1}{2x+y}\ge\dfrac{4}{x+2y+2x+y}=\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3\left(x+y\right)}=4\)
Dấu '' = "xảy ra khi và chỉ khi x + 2y = y + 2x (=) x=y
Áp dụng BĐT Cauchy - Schwarz dạng Engel ta có :
\(\dfrac{a^2}{a-1}+\dfrac{b^2}{b-1}\ge\dfrac{\left(a+b\right)^2}{a+b-2}=\dfrac{\left(a+b\right)^2-4+4}{a+b-2}\)
\(=\dfrac{\left(a+b+2\right)\left(a+b-2\right)+4}{a+b-2}=a+b+2+\dfrac{4}{a+b-2}\)
\(=a+b-2+\dfrac{4}{a+b-2}+4\ge2\sqrt{\left(a+b-2\right).\dfrac{4}{a+b-2}}-4=0\)
Cách của bạn đúng rồi nhưng bạn nhầm dấu ở dòng cuối, phải là +4. Min=8 khi x=y=2