Useful Substitutions in Inequalities

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 $\dfrac{1}{ab+b}+\dfrac{1}{bc+c}+\dfrac{1}{ca+a}\ge \dfrac{3}{2}.$

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We apply the substitution $a=\dfrac{x}{y}, b=\dfrac{y}{z}, c=\dfrac{z}{x}$ to get rid of the condition $abc=1$. The inequality becomes $\dfrac{1}{\dfrac{x}{y}\cdot \dfrac{y}{z}+\dfrac{y}{z}}+\dfrac{1}{\dfrac{y}{z}\cdot \dfrac{z}{x}+\dfrac{z}{x}}+\dfrac{1}{\dfrac{z}{x}\cdot \dfrac{x}{y}+\dfrac{x}{y}}=\dfrac{z}{x+y}+\dfrac{x}{y+z}+\dfrac{y}{x+z} \ge \dfrac{3}{2}.$ However, the last inequality is simply Nesbitt's inequality, so we are done. $_\square$

Given that $a,b,c>0$, prove that $\sqrt{\dfrac{a+b}{c}+2}+\sqrt{\dfrac{b+c}{a}+2}+\sqrt{\dfrac{c+a}{b}+2}\ge 6.$

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We see that $\sqrt{\dfrac{a+b}{c}+2}+\sqrt{\dfrac{b+c}{a}+2}+\sqrt{\dfrac{c+a}{b}+2}=\sqrt{\dfrac{a+b+c}{c}+1}+\sqrt{\dfrac{a+b+c}{a}+1}+\sqrt{\dfrac{a+b+c}{b}+1}.$ We apply the substitution $x=\dfrac{a}{a+b+c}, y=\dfrac{b}{a+b+c}, z=\dfrac{c}{a+b+c}$ to get the condition $x+y+z=1$ and the inequality as $\sqrt{\dfrac{1}{x}+1}+\sqrt{\dfrac{1}{y}+1}+\sqrt{\dfrac{1}{z}+1}\ge 6.$ Now notice that the function $f(x)=\sqrt{\dfrac{1}{x}+1}$ is convex, so by Jensen's inequality $\sqrt{\dfrac{1}{x}+1}+\sqrt{\dfrac{1}{y}+1}+\sqrt{\dfrac{1}{z}+1} \ge 3\sqrt{\dfrac{1}{\frac{x+y+z}{3}}+1} = 6,$ implying that we are done. $_\square$

Given that $a,b,c>0$ are reals satisfying $a+b+c=abc$, prove that $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\ge \dfrac{3}{4}.$

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Utilizing the substitution $a=\tan A, b=\tan B, c=\tan C$ with $A+B+C=\pi$ rids the condition and turns the inequality to $\begin{aligned} \dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1} &= \dfrac{1}{\tan^2 A+1}+\dfrac{1}{\tan^2 B+1}+\dfrac{1}{\tan^2C+1} \\&= \cos^2A+\cos^2B+\cos^2C \\&\ge \dfrac{3}{4}, \end{aligned}$ where the identity $\cos^2A+\cos^2B+\cos^2C =1-2\cos A\cos B\cos C$ and the well-known inequality $\cos A\cos B\cos C \le \dfrac{1}{8}$ were used. $_\square$

Given that $a,b,c>0$ such that $ab+bc+ca=1$, prove that $\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge 3+\sqrt{1+\dfrac{1}{a^2}}+\sqrt{1+\dfrac{1}{b^2}}+\sqrt{1+\dfrac{1}{c^2}}.$

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Using the substitution $a=\cot A, b=\cot B, c=\cot C$ with $A+B+C=\pi$ rids the condition and reduces the inequality to $\begin{aligned} LHS&= \tan A\tan B+\tan B\tan C+\tan C\tan A\\ &=\dfrac{\sin A\sin B\cos C+\sin A\cos B\sin C+\cos A\sin B\sin C}{\cos A\cos B\cos C}\\ &= \dfrac{\cos A\cos B\cos C -\cos (A+B+C)}{\cos A\cos B\cos C}\\ &= 1+\dfrac{1}{\cos A\cos B\cos C}\\ \\ RHS &= 3+\sqrt{1+\tan^2 A}+\sqrt{1+\tan^2 B}+\sqrt{1+\tan^2 C}\\ &= 3+\dfrac{1}{\cos A}+\dfrac{1}{\cos B}+\dfrac{1}{\cos C}. \end{aligned}$ It remains to prove $\dfrac{1}{\cos A\cos B\cos C}\ge 2+\dfrac{1}{\cos A}+\dfrac{1}{\cos B}+\dfrac{1}{\cos C}. \ _\square$

If $A+B+C=\pi$, prove $\dfrac{1}{\cos A\cos B\cos C}\ge 2+\dfrac{1}{\cos A}+\dfrac{1}{\cos B}+\dfrac{1}{\cos C}$

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We utilize the substitution $x=\cos A, y=\cos B, z=\cos C$ to get the condition $x^2+y^2+z^2+2xyz=1$ and the inequality reduces to $\dfrac{1}{xyz}\ge 2+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ However, dividing both sides of the condition by $xyz$ gives $\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}+2=\dfrac{1}{xyz}$ and substituting this into our inequality, it remains to prove $\dfrac{x}{yz}+\dfrac{y}{zx}+\dfrac{z}{xy}\ge \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ However, this is true as by AM-GM as $\dfrac{x}{yz}+\dfrac{y}{zx}\ge 2\sqrt{\dfrac{x}{yz}\cdot \dfrac{y}{zx}}=\dfrac{2}{z^2}$, and summing this inequality cyclically gives the inequality we want to prove. $_\square$

1) Given that $a,b,c>0$ are reals satisfying $a+b+c+2=abc$, prove that $a+b+c\ge \dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}$ 2) Given that $a,b,c >0$ are reals satisfying $ab+bc+ca+2abc=1$, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 4(a+b+c)$

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These two problems, if you didn't notice, are intricately connected: in fact, one is just the other with the substitution of variables $(a,b,c)\mapsto \left(\dfrac{1}{a}, \dfrac{1}{b},\dfrac{1}{c}\right)$.

With this in mind, the substitution for the the condition $a+b+c+2=abc$ is $a=\dfrac{x+y}{z}, b=\dfrac{y+z}{x}, c=\dfrac{z+x}{y}$ and the substitution for the condition $ab+bc+ca+2abc=1$ is therefore $a=\dfrac{x}{y+z}, b=\dfrac{y}{z+x}, c=\dfrac{z}{x+y}$.

Now with this substitution both problems become the inequality $\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}\ge \dfrac{4x}{y+z}+\dfrac{4y}{z+x}+\dfrac{4z}{x+y}$ However, $\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{z+x}{y}=\left(\dfrac{x}{y}+\dfrac{x}{z}\right)+\left(\dfrac{y}{z}+\dfrac{y}{x}\right)+\left(\dfrac{z}{y}+\dfrac{z}{x}\right) \ge \dfrac{4x}{y+z}+\dfrac{4y}{z+x}+\dfrac{4z}{x+y}$ by the inequality $\dfrac{1}{x}+\dfrac{1}{y}\ge \dfrac{4}{x+y}$, so we are done. $_\square$

Prove that if $a,b,c\ge 0$ satisfy the condition $|a^2+b^2+c^2-4|=abc$, then $(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)\ge 0$

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Before we solve this problem, some background on the necessary substitution is needed. Given the condition $a^2+b^2+c^2-abc=4$ and $\text{max}\{a,b,c\} \ge 2$, we can rid the condition with the substitution $a=x+\dfrac{1}{x}, b=y+\dfrac{1}{y}, c=z+\dfrac{1}{z}$ such that $xyz=1$.

Now back to the problem: we see that if $\text{max}\{a,b,c\} < 2$, then clearly the inequality is true. Otherwise, $\text{max}\{a,b,c\} \ge 2$. This means that $a^2+b^2+c^2-4 \ge 0$ so the condition simply turns into $a^2+b^2+c^2-4=abc\implies a^2+b^2+c^2-abc=4$ Thus, we can use the substitution described above. However, this means that $a=x+\dfrac{1}{x}\ge 2$ by AM-GM; similarly, $b,c\ge 2$ so the inequality we seek is true again. $_\square$