Cho 2(a + 1)(b + 1) = (a + b)(a + b +2). CMR: \(a^2\)+ \(b^2\) = 2
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 = 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+4b+1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 +1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 +2009/ab+bc+ac >=670
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 = 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+4b+1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 +1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 +2009/ab+bc+ac >=670
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 = 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+4b+1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 +1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 +2009/ab+bc+ac >=670
Bất đẳng thức Bunhiacopxki
B1: Cho a,b,c thỏa mãn: a+b+c=1. CMR: \(a^2+b^2+c^2\ge\dfrac{1}{3}\)
B2: Cho a,b,c dương thỏa mãn: \(a^2+4b^2+9c^2=2015\). CMR: \(a+b+c\le\dfrac{\sqrt{14}}{6}\)
B3: Cho a,b dương thỏa mãn: \(a^2+b^2=1\).CMR: \(a\sqrt{1+a}+b\sqrt{1+b}\le\sqrt{2+\sqrt{2}}\)
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$
$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$
$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Bài 2:
Áp dụng BĐT Bunhiacopxky:
$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$
$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$
$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$
$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>
Bài 3:
Áp dụng BĐT Bunhiacopxky:
$2=(a^2+b^2)(1+1)\geq (a+b)^2\Rightarrow a+b\leq \sqrt{2}$
$(a\sqrt{1+a}+b\sqrt{1+b})^2\leq (a^2+b^2)(1+a+1+b)$
$=2+a+b\leq 2+\sqrt{2}$
$\Rightarrow a\sqrt{1+a}+b\sqrt{1+b}\leq \sqrt{2+\sqrt{2}}$
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{\sqrt{2}}$
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 + 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+ 4b + 1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 + 1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 + 2009/ab+bc+ac >=670
B1: Cho \(0\le a,b,c\le2\) thỏa mãn \(a+b+c=3\). CMR: \(a^2+b^2+c^2\le5\)
B2: Cho \(a,b\ge0\) thỏa mãn \(a^2+b^2=a+b\). TÌm GTLN \(S=\dfrac{a}{a+1}+\dfrac{b}{b+1}\)
B3: CMR: \(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\forall x\ne y,xy\ne0\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2:
Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:
\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{(a+1)(b+1)(c+1)}{abc}\)
\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)
\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
1) Cho a, b, c ∈ [0;1] và a + b + c = 2. CMR ab + bc + ca ≥ 2abc + \(\dfrac{20}{27}\)
2) Cho a, b, c ∈ [1;3] và a + b + c = 6. CMR a3 + b3 + c3 ≤ 36
3) Cho các số dương a, b, c, d thoả mãn a + b + c + d = 4. CMR \(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+d^2}+\dfrac{d}{1+a^2}\) ≥ 2
1.
Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có 2 số cùng phía so với \(\dfrac{2}{3}\), không mất tính tổng quát, giả sử đó là b và c
\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\ge0\)
Mặt khác \(0\le a\le1\Rightarrow1-a\ge0\)
\(\Rightarrow\left(b-\dfrac{2}{3}\right)\left(c-\dfrac{2}{3}\right)\left(1-a\right)\ge0\)
\(\Leftrightarrow-abc\ge\dfrac{4a}{9}+\dfrac{2b}{3}+\dfrac{2c}{3}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}\)
\(\Leftrightarrow-abc\ge-\dfrac{2a}{9}+\dfrac{2}{3}\left(a+b+c\right)-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc-\dfrac{4}{9}=-\dfrac{2a}{9}-\dfrac{2ab}{3}-\dfrac{2ac}{3}-bc+\dfrac{8}{9}\)
\(\Leftrightarrow-2abc\ge-\dfrac{4a}{9}-\dfrac{4ab}{3}-\dfrac{4ac}{3}-2bc+\dfrac{16}{9}\)
\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{ab}{3}-\dfrac{ac}{3}-bc+\dfrac{16}{9}\)
\(\Leftrightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(b+c\right)-bc+\dfrac{16}{9}\ge-\dfrac{4a}{9}-\dfrac{a}{3}\left(2-a\right)-\dfrac{\left(b+c\right)^2}{4}+\dfrac{16}{9}\)
\(\Rightarrow ab+bc+ca-2abc\ge-\dfrac{4a}{9}+\dfrac{a^2}{3}-\dfrac{2a}{3}-\dfrac{\left(2-a\right)^2}{4}+\dfrac{16}{9}\)
\(\Rightarrow ab+bc+ca-2abc\ge\dfrac{a^2}{12}-\dfrac{a}{9}+\dfrac{7}{9}=\dfrac{1}{12}\left(a-\dfrac{2}{3}\right)^2+\dfrac{20}{27}\ge\dfrac{20}{27}\)
\(\Rightarrow ab+bc+ca\ge2abc+\dfrac{20}{27}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
2.
Đặt \(\left(a;b;c\right)=\left(x+1;y+1;z+1\right)\Rightarrow\left\{{}\begin{matrix}x;y;z\in\left[0;2\right]\\x+y+z=3\end{matrix}\right.\)
Ta có: \(P=\left(x+1\right)^3+\left(y+1\right)^3+\left(z+1\right)^3\)
\(P=x^3+y^3+z^3+3\left(x^2+y^2+z^2\right)+12\)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\Rightarrow x\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}y^3+z^3=\left(y+z\right)^3-3yz\left(y+z\right)\le\left(y+z\right)^3\\y^2+z^2=\left(y+z\right)^2-2yz\le\left(y+z\right)^2\end{matrix}\right.\)
\(\Rightarrow P\le x^3+\left(3-x\right)^3+3x^2+3\left(3-x\right)^2+12\)
\(\Rightarrow P\le15x^2-45x+66=15\left(x-1\right)\left(x-2\right)+36\le36\)
(Do \(1\le x\le2\Rightarrow\left(x-1\right)\left(x-2\right)\le0\))
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(2;1;0\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(1;2;3\right)\) và các hoán vị
a, Cho a+b+c=0 CMR:\(a^3\)+\(a^2c-abc+b^2c+b^3=0\)
b, Cho 2(a+1)(b+1)=(a+b)(a+b+2) CMR:\(a^2+b^2=2\)
c, Cho \(a^2+c^2=2b^2\)CMR;
(a+b)(a+c)+(c+a)(c+b)=2(b+a)(b+c)
a. \(a^3+a^2c-abc+b^2c+b^3\)
<=> \(\left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)\)
<=> (\(\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
<=> \(\left(a+b+c\right)\left(a^2-ab+b^2\right)\)
vì a+b+c =0 => đpcm
b. 2(a+1)(b+1)=(a+b)(a+b+2)
<=> \(2\left(ab+a+b+1\right)=\)\(a^2+ab+2a+ab+b^2+2b\)
<=> \(2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b\)
<=> \(a^2+b^2=2\)=> đpcm
a. a^3+a^2c-abc+b^2c+b^3a3+a2c−abc+b2c+b3
<=> \left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)(a3+b3)+c(a2−ab+b2)
<=> (\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)(a+b)(a2−ab+b2)+c(a2−ab+b2)
<=> \left(a+b+c\right)\left(a^2-ab+b^2\right)(a+b+c)(a2−ab+b2)
vì a+b+c =0 => đpcm
b. 2(a+1)(b+1)=(a+b)(a+b+2)
<=> 2\left(ab+a+b+1\right)=2(ab+a+b+1)=a^2+ab+2a+ab+b^2+2ba2+ab+2a+ab+b2+2b
<=> 2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b2ab+2a+2b+2=a2ab+2a+ab+b2+2b
<=> a^2+b^2=2a2+b2=2=> đpcm
cho \( a+b+c=abc\)
CMR \(a(b^2-1)(c^2-1)+b(a^2-1)(c^2-1)+c(a^2-1)(b^2-1)\)
\(=4abc\)
\(a\left(b^2-1\right)\left(c^2-1\right)+b\left(a^2-1\right)\left(c^2-1\right)+c\left(a^2-1\right)\left(b^2-1\right)\\ =\left(ab^2-a\right)\left(c^2-1\right)+\left(a^2b-b\right)\left(c^2-1\right)+\left(a^2c-c\right)\left(b^2-1\right)\\ =ab^2c^2-ab^2-ac^2+a+a^2bc^2-a^2b-bc^2+b+a^2b^2c-a^2c-b^2c+c\\ =abc\left(ab+bc+ac\right)-\left(a^2b+ab^2+ac^2+bc^2+a^2c+b^2c\right)+\left(a+b+c\right)\\ =abc\left(ab+bc+ca\right)+\left(a+b+c\right)+3abc-\left[\left(a^2b+ab^2+abc\right)+\left(b^2c+bc^2+abc\right)+\left(a^2c+ac^2+abc\right)\right]\\ =abc\left(ab+bc+ca\right)+abc+3abc-\left[ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+b+c\right)\right]\\ =4abc+abc\left(ab+bc+ca\right)-\left(a+b+c\right)\left(ab+bc+ca\right)\\ =4abc+abc\left(ab+bc+ca\right)-abc\left(ab+bc+ca\right)=4abc\)