How to solve: b^2=ac than prove that (ab+bc+ca)^3=abc(a+b+c)^3?
Work on the left hand side. Replace ac by b^2. (ab + bc + ac)^3 = (ab + bc + b^2)^3 Now take b out as common factor [ b*(a + c + b) ]^3 Now cube each part inside separately (b^3)(a + b + c)^3 Replace b^2 of the b^3 by ac to get abc(a + b + c)^3 which is the right hand side.
If (a+b+c)^3 = 27abc then prove that a=b=c?
Let a=A³, b=B³, c=C³ then given equality becomes A³+B³+C³−3ABC=0 But (A+B+C)³ ≡ A³+B³+C³ + 3(A+B+C)(AB+BC+CA) − 3ABC ∴ (A+B+C)(A²+B²+C²−AB−BC−CA) = ½(A+B+C)( (A−B)²+(B−C)²+(C−A)² ) = 0 It follows that A³+B³+C³−3ABC = 0 <==> A+B+C=0 or (A−B)²+(B−C)²+(C−A)² = 0 So given equality is true if A+B+C=0 or A=B=C … (i) Only if a,b,c are non-negative do both options in (i) imply a=b=c If a,b,c are unrestricted then given equality can be true if A+B+C=0 In terms of a,b,c this is ³√a + ³√b + ³√c = 0 which does not imply a=b=c e.g. a=8, b=−27, c=1, a+b+c=−18, abc=−216, (a+b+c)³=27abc with a,b,c unequal @pinkgreen : Why does 27a³+27(x1+x2)a²+9(x1+x2)²a+(x1+x2)³ = 27a³+27(x1+x2)a²+27(x1)(x2)a => x1+x2=3(x1)(x2) and (x1+x2)³=0 ?
If a + b + c=5 and ab + bc + ca = 10 then prove that a^3 + b^3 + c^3 – 3abc = -25?
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Prove that a^2b^2+b^2c^2+c^2a^2>=abc(a+b+... ?
Are you testing us? Haha, okay. Move everything to one side, and it becomes the task of proving the inequality: a^2 (b^2 - bc) + b^2 (c^2 - ca) + c^2 (a^2 - ab) >= 0. Let this equation be known as (1). Trivially, the inequality is true if a = b = c. Otherwise, assume that a >= b >= c > 0. Then the first term is obviously positive. The second term, however, is not because c^2 - ca < 0. This is the problematic term. We move on to the third term c^2 (a^2 - ab). This term is positive as well. Now, we complete all of the incomplete squares by adding in a^2 (-bc + c^2) + b^2 (-ca + a^2) + c^2 (-ab + b^2). To counteract, we add a^2 bc - a^2c^2 + b^2 ca - b^2a^2 + c^2 ab - c^2 b^2 to the same side. The complete-square section is a^2(b - c)^2 + b^2(c - a)^2 + c^2(a - b)^2 and the added part is a^2 bc + b^2 ca + c^2 ab - a^2c^2 - b^2c^2 - c^2a^2. However, notice that the added part is EXACTLY the same as the equation (1). And since the completed square section is ALWAYS positive, half of something positive is also positive and so follows our result. Hope I surprised you. ;)
If a, b, c are positive and not all equal, how do I derive (a + b + c)(ab + bc + ca) > 9abc and (b + c)(c + a)(a + b) > 8abc?
Consider 3 positive numbers a,b and cUsing Arithmetic–geometric mean Inequality we can write[math]\frac{a+b+c}{3}\geqslant \sqrt[3]{a.b.c}[/math] With equality holding true only if a = b = c. As per the given condition [math]a \neq b \neq c [/math]we can write[math]\frac{a+b+c}{3} > \sqrt[3]{a.b.c}[/math]Similarly consider 3 positive numbers ab,bc and caAgain using Arithmetic–geometric mean Inequality we can write[math]\frac{ab+bc+ca}{3}\geqslant \sqrt[3]{a^2.b^2.c^2}[/math]With equality holding true only if ab = bc = ca. As per the given condition we can write[math]\frac{ab+bc+ca}{3} > \sqrt[3]{a^2.b^2.c^2}[/math]As all the terms in both inequalities are positive . Hence we may multiply the above 2 inequalities We get [math]\frac{(a+b+c)(ab+bc+ca)}{(3).(3)} > \sqrt[3]{a.b.c}. ^3\sqrt{a^2.b^2.c^2}[/math][math]\implies \frac{(a+b+c)(ab+bc+ca)}{9} > \sqrt[3]{a^3.b^3.c^3} [/math][math]\implies \frac {(a+b+c)(ab+bc+ca)}{9} > a.b.c [/math] [math]\implies (a+b+c)(ab+bc+ca) > 9.a.b.c [/math]Hence provedUsing a similar approach in the second case Consider 2 positive numbers b,cUsing Arithmetic–geometric mean Inequality we can write[math]\frac {b+c}{2}\geqslant \sqrt{b.c}[/math]With similarity at b = c .As per the given condition we can write [math]\frac {b+c}{2} > \sqrt{b.c}[/math]Similarly for a,c and a,b we can get the inequalities [math]\frac {b+a}{2} > \sqrt{b.a}[/math][math]\frac {a+c}{2} > \sqrt{a.c}[/math] As all the terms in both inequaltites are positive we may multiply the above 3 inequalities we get [math]\frac {(b+c).(b+a).(a+c)}{2.2.2} > \sqrt{b.c}\sqrt{b.a}\sqrt{a.c}[/math][math]\frac {(b+c).(b+a).(a+c)}{8} > \sqrt{b.^2c^2.a^2}[/math][math]\frac {(b+c).(b+a).(a+c)}{8} > b.c.a[/math][math](b+c).(b+a).(a+c) > 8.b.c.a[/math]Hence proved
How does one prove that [math]a^3+b^3+c^3=3abc[/math]?
Others have already pointed out that the given equation is not always true, but I’m wondering if the OP meant to ask about [math]a^3 + b^3 + c^3 \geq 3abc[/math]. This inequality is true for all non-negative real numbers [math]a,b,c[/math] and it follows immedaitely from the AM-GM inequality (a proof of the general case can be found here but I’ll just prove it for three variables.)We have[math]a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc) \tag*{}[/math]If [math]a,b,c \geq 0[/math] then the left-most bracket is non-negative. To prove the right-most bracket is non-negative, we need to use the AM-GM inequality in two variables. Specifically, it may be written as[math]\frac{1}{2}((a-b)^2+(a-c)^2+(b-c)^2) \tag*{}[/math]and this is clearly non-negative. Hence, [math]a^3+b^3+c^3 -3abc \geq 0[/math] and the result follows.
A+b+c=6,ab+bc+ca=11,abc=6,then what is (1+a)(1+b)(1+c)=?
OK, so there is a nice trick for getting the answer really fast. Just for the heck of it, let's do it the hard way. a + b + c = 6 ab + bc + ca = 11 abc = 6 First, note that swapping any two variables, say a and b, results in the same three equations. Such a set of equations is said to be symmetric in a, b, and c. We have three equations and three unknowns, with luck we will get three answers when we solve for one of the variables. If we do, then, because of symmetry, those three answers will be a, b, and c in whatever order we choose them. So let's solve for c. (1) From a + b + c = 6, we get a + b = 6 - c (2) Multiply through by c and you get ac + bc = 6c - c^2 (3) From ab + bc + ca = 11, we get ac + bc = 11 - ab. (4) Substituting (3) into (2), we get 11 - ab = 6c - c^2 (5) Multiply through by c, and we get 11c - abc = 6c^2 - c^3 (6) But abc = 6. So 11c - 6 = 6c^2 - c^3 Rearranging, we get c^3 - 6c^2 + 11c - 6 = 0 (c^3 - 3c^2 + 2c) + (-3c^2 + 9c - 6) = 0 c(c^2 - 3c + 2) - 3(c^2 - 3c + 2) = 0 (c - 3)(c^2 - 3c + 2) = 0 (c - 3)(c - 2)(c - 1) = 0 So a, b, and c equals 1, 2, and 3 in whatever order you want. Hence (1 +a)(1 + b)(1 + c) = (2)(3)(4) = 24
If abc+bc+ab+a+b+c=1000 then what is a+b+c=? (a,b, c are positive integers)
I guess there is typo in the question, the equation should be a symmetrical one and read,[math]\large{abc\>+\>ab\>+\>bc\>+\>ca\>+\>a\>+\>b\>+\>c\>=\>1000}[/math]then what is the value of [math]\large{a+b+c}[/math]The solution to this problem begins by adding 1 to both sides of the equation.So,[math]\large{abc\>+\>ab\>+\>bc\>+\>ca\>+\>a\>+\>b\>+\>c\>+\>1\>=\>1000\>+\>1}[/math]This can be reduced to,[math]\large{ab.(c\>+\>1)\>+\>b.(c\>+\>1)\>+\>a.(c\>+\>1)\>+\>(c\>+\>1)\>=\>1001}[/math]This can be further reduced to,[math]\large{(c\>+\>1).(ab\>+\>b\>+\>a\>+\>1)=\>1001}[/math]Which reduces to,[math]\large{(a\>+\>1).(b\>+\>1).(c\>+\>1)\>=\>7\>\times\>11\>\times\>13}[/math]Since exact values of [math]\large{a,b,c} [/math] are not required, so order of factors is not important, so in any order,[math]\large{a\>=\>7\>-\>1\>=\>6}[/math][math]\large{b\>=\>11\>-\>1\>=\>10}[/math][math]\large{a\>=\>13\>-\>1\>=\>12}[/math][math]\Large{\therefore a\>+\>b\>+\>c\>=\>6\>+\>10\>+\>12\>=\>\boldsymbol{28}}[/math]
If a , b, c are unequal and positive real numbers then prove that ...help please ?
as a results of fact the 4 numbers are in a geometrical progression, they are in a position to be written as a, ar, ar^2, ar^3. a is believed to be advantageous and so is r=b/a. you opt for to instruct that a+d?b+c, i.e. that d-c-b+a?0, or that a(r^3-r^2-r+a million)?0. yet a>0 and r^3-r^2-r+a million = (r^2-a million)(r-a million) = (r+a million)·(r-a million)^2, the place r+a million>0 and (r-a million)^2, being a sq., is likewise advantageous till r=a million (wherein case a=b=c=d). So the product is surely ?0!
If 1/a+1/b+1/c=1/a+b+c , then how do you prove that 1/a5+1/b5+1/c5=1/a5+b5+c5?
Given,[math]\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}[/math][math]\implies \dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{a+b+c}-\dfrac{1}{c}[/math][math]\implies \dfrac{a+b}{ab}=\dfrac{c-(a+b+c)}{c(a+b+c)}[/math][math]\implies \dfrac{a+b}{ab}=\dfrac{-(a+b)}{c(a+b+c)}[/math]Conclusions:Case 1:[math](a+b)=0[/math][math]\implies b=-a[/math][You may also get [math](b+c)=0[/math] or [math](c+a)=0[/math] depending on how we group the terms.]Case 2:[math](a+b) \ne 0[/math]We can continue we our earlier calculation.[math]\dfrac{a+b}{ab}=\dfrac{-(a+b)}{c(a+b+c)}[/math][math]\implies \dfrac{1}{ab}=\dfrac{-1}{c(a+b+c)}[/math][math]\implies \dfrac{1}{ab}+\dfrac{1}{c(a+b+c)}=0[/math][math]\implies c(a+b+c)+ab=0[/math][math]\implies c^2=-(ab+bc+ca)[/math]It can shown that[math]a^2=-(ab+bc+ca)[/math][math]b^2=-(ab+bc+ca)[/math][math]c^2=-(ab+bc+ca)[/math]This implies [math]a^2=b^2=c^2=k^2[/math](where 'k' is a positive constant.[math]a=\pm k[/math][math]b=\pm k[/math][math]c=\pm k[/math]From[math]\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}[/math]We can see that possible values of (a,b,c) can be[math](k,k,-k) \text{ or } (k,-k,k) \text{ or } (-k,k,k) \text{ or } (-k,-k,k) \text{ or } (-k,k,-k) \text{ or } (k,-k,-k)[/math]This just means that 2 of (a,b,c) will be equal and the other 1 of (a,b,c) will be opposite in sign.We will just consider the case where[math](a,b,c)\equiv (k,k,-k).[/math]Required Proof:Case 1:L.H.S:[math]\dfrac{1}{a^5}+\dfrac{1}{b^5}+\dfrac{1}{c^5}=\dfrac{1}{a^5}+\dfrac{1}{(-a)^5}+\dfrac{1}{c^5}=\dfrac{1}{c^5}[/math]R.H.S[math]\dfrac{1}{a^5+b^5+c^5}=\dfrac{1}{a^5+(-a)^5+c^5}=\dfrac{1}{c^5}[/math]L.H.S=R.H.SCase 2:[math]a=k,b=k,c=-k[/math]L.H.S.[math]\dfrac{1}{a^5}+\dfrac{1}{b^5}+\dfrac{1}{c^5}=\dfrac{1}{k^5}+\dfrac{1}{k^5}+\dfrac{1}{(-k)^5}=\dfrac{1}{k^5}[/math]R.H.S.[math]\dfrac{1}{a^5+b^5+c^5}=\dfrac{1}{k^5+k^5+(-k)^5}=\dfrac{1}{k^5}[/math]L.H.S.=R.H.S.This completes the proof.