Inequality form olympiad

Question

Let a , b , c be real numbers such that $a^2+b^2+c^2=2$

Show that $a+b+c \leq abc+2$

Please help. I'm stuck on this one. I tried a lot of things. The problem is I can't use standard inequalities because they're not necessarily positive.

Answer 1

Proof 1. Using the Cauchy-Schwarz inequality, we have $$[a(1-bc) +(b+c)]^2 \leqslant [a^2+(b+c)^2][(1-bc)^2+1] = 2(1+bc)[(1-bc)^2+1].$$ Thefore, we need to prove $$(1+bc)[(1-bc)^2+1] \leqslant 2,$$ or $$b^2c^2(bc-1) \leqslant 0.$$ Which is true because $$bc \leqslant \frac{b^2+c^2}{2} \leqslant \frac{a^2+b^2+c^2}{2}=1.$$ Done.

Proof 2. We need to prove $$4-(a+b+c-abc)^2 \geqslant 0.$$ equivalent to $$4-(a+b+c)^2+2abc(a+b+c)-4a^2b^2c^2 \geqslant 0.$$ Setting $A=a^2+b^2+c^2=2,$ the inequality equivalent to $$4[4-(a+b+c)^2+2abc(a+b+c)-a^2b^2c^2] \ge 0$$ or $$2^3-2^2 \cdot [(a+b+c)^2-2]+2 \cdot 4abc \cdot (a+b+c)-4a^2b^2c^2\geqslant 0,$$ $$A^3-A^2\cdot\left[(a+b+c)^2-A\right]+4 \cdot A \cdot abc \cdot (a+b+c)-4a^2b^2c^2 \geqslant 0,$$ or $$(2-2ab)(2-2bc)(2-2ca)+4a^2b^2c^2 \geqslant 0.$$ Which is true because $$2-2ab=(a^2-2ab+b^2)+c^2=(a-b)^2+c^2 \geqslant 0.$$ The proof is completed.