Sometimes when faced with an inequality, there is an inconvenient condition on the variables that makes everything seem much harder. How do we get rid of the said condition?
In fact, knowing some clever substitutions will often rid those conditions and give us a conditionless inequality, something we all would prefer.
Other times, we are faced with complicated inequality that we're not quite sure how to tackle. In these cases, it's worth it to actually introduce a condition with a substitution.
In either case, clever substitutions will help us immensely.
Given that a,b,c>0 satisfy abc=1, prove that ab+b1+bc+c1+ca+a1≥23.
We apply the substitution a=yx,b=zy,c=xz to get rid of the condition abc=1. The inequality becomes yx⋅zy+zy1+zy⋅xz+xz1+xz⋅yx+yx1=x+yz+y+zx+x+zy≥23.
However, the last inequality is simply Nesbitt's inequality, so we are done.
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Given that a,b,c>0, prove that ca+b+2+ab+c+2+bc+a+2≥6.
We see that ca+b+2+ab+c+2+bc+a+2=ca+b+c+1+aa+b+c+1+ba+b+c+1. We apply the substitution x=a+b+ca,y=a+b+cb,z=a+b+cc to get the condition x+y+z=1 and the inequality as x1+1+y1+1+z1+1≥6. Now notice that the function f(x)=x1+1 is convex, so by Jensen's inequalityx1+1+y1+1+z1+1≥33x+y+z1+1=6, implying that we are done. □
Given that a,b,c>0 are reals satisfying a+b+c=abc, prove that a2+11+b2+11+c2+11≥43.
Utilizing the substitution a=tanA,b=tanB,c=tanC with A+B+C=π rids the condition and turns the inequality to
a2+11+b2+11+c2+11=tan2A+11+tan2B+11+tan2C+11=cos2A+cos2B+cos2C≥43,
where the identity cos2A+cos2B+cos2C=1−2cosAcosBcosC and the well-known inequality cosAcosBcosC≤81 were used. □
Given that a,b,c>0 such that ab+bc+ca=1, prove that ab1+bc1+ca1≥3+1+a21+1+b21+1+c21.
Using the substitution a=cotA,b=cotB,c=cotC with A+B+C=π rids the condition and reduces the inequality to
LHSRHS=tanAtanB+tanBtanC+tanCtanA=cosAcosBcosCsinAsinBcosC+sinAcosBsinC+cosAsinBsinC=cosAcosBcosCcosAcosBcosC−cos(A+B+C)=1+cosAcosBcosC1=3+1+tan2A+1+tan2B+1+tan2C=3+cosA1+cosB1+cosC1.
It remains to prove cosAcosBcosC1≥2+cosA1+cosB1+cosC1.□
If A+B+C=π, prove cosAcosBcosC1≥2+cosA1+cosB1+cosC1
We utilize the substitution x=cosA,y=cosB,z=cosC to get the condition x2+y2+z2+2xyz=1 and the inequality reduces to xyz1≥2+x1+y1+z1
However, dividing both sides of the condition by xyz gives yzx+zxy+xyz+2=xyz1 and substituting this into our inequality, it remains to prove yzx+zxy+xyz≥x1+y1+z1
However, this is true as by AM-GM as yzx+zxy≥2yzx⋅zxy=z22, and summing this inequality cyclically gives the inequality we want to prove. □
1) Given that a,b,c>0 are reals satisfying a+b+c+2=abc, prove that a+b+c≥a4+b4+c4
2) Given that a,b,c>0 are reals satisfying ab+bc+ca+2abc=1, prove that a1+b1+c1≥4(a+b+c)
These two problems, if you didn't notice, are intricately connected: in fact, one is just the other with the substitution of variables (a,b,c)↦(a1,b1,c1).
With this in mind, the substitution for the the condition a+b+c+2=abc is a=zx+y,b=xy+z,c=yz+x and the substitution for the condition ab+bc+ca+2abc=1 is therefore a=y+zx,b=z+xy,c=x+yz.
Now with this substitution both problems become the inequality zx+y+xy+z+yz+x≥y+z4x+z+x4y+x+y4z
However, zx+y+xy+z+yz+x=(yx+zx)+(zy+xy)+(yz+xz)≥y+z4x+z+x4y+x+y4z by the inequality x1+y1≥x+y4, so we are done. □
Prove that if a,b,c≥0 satisfy the condition ∣a2+b2+c2−4∣=abc, then (a−2)(b−2)+(b−2)(c−2)+(c−2)(a−2)≥0
Before we solve this problem, some background on the necessary substitution is needed. Given the condition a2+b2+c2−abc=4andmax{a,b,c}≥2, we can rid the condition with the substitution a=x+x1,b=y+y1,c=z+z1 such that xyz=1.
Now back to the problem: we see that if max{a,b,c}<2, then clearly the inequality is true. Otherwise, max{a,b,c}≥2. This means that a2+b2+c2−4≥0 so the condition simply turns into a2+b2+c2−4=abc⟹a2+b2+c2−abc=4
Thus, we can use the substitution described above. However, this means that a=x+x1≥2 by AM-GM; similarly, b,c≥2 so the inequality we seek is true again. □