pt <=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c}{a+b+c}=5\) (Cộng 4 vào mỗi vế)
<=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c-5\left(a+b+c\right)}{a+b+c}=0\)
<=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4a-4b-4c}{a+b+c}=0\)
<=> \(\left(a+b+c-x\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\right)=0\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng engel, ta có :
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}>\frac{4}{a+b+c}\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}>0\)
Vậy phương trình trên có nghiệm là
x = a + b + c
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