triangle112 wrote: If $0< a,b,c<1$ prove that $a+b+c+2abc>ab+bc+ca+2\sqrt{abc}$ $\text{\LaTeX}$ed @below done

Thats it. Have fun

triangle112 wrote: If a b c positive reals smaller than 1, prove: a+b+c+2abc>ab+bc+ca+2(abc)^(1/2) Let $a=\cos^2\alpha$ and $b=\cos^2\beta$, where $\{\alpha,\beta\}\subset(0,90^{\circ}).$ Thus, we need to prove that $$1+abc-2\sqrt{abc}>1-a-b-c+ab+ac+bc-abc$$or $$(1-\sqrt{abc})^2>(1-a)(1-b)(1-c)$$or $$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<1,$$which is true because $$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt{ab}+\sqrt{(1-a)(1-b)}=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta)\leq1.$$

arqady wrote: $$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<1,$$which is true because $$\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt{ab}+\sqrt{(1-a)(1-b)}=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta)\leq1.$$ Can be done also: if $x\in (0,1)$ then $x^{1/2}<x^{1/3}$. So we know $\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<\sqrt[3]{abc}+\sqrt[3]{(1-a)(1-b)(1-c)}\le \frac{a+b+c}{3}+\frac{(1-a)+(1-b)+(1-c)}{3}=1$ by AG.

Let $a,b,c\in (0,1). $ Prove that $$a+b+c+2abc > ab+bc+ca+2\sqrt{abc}.$$JBMO TST Serbia 2019

Let $x,y,z\in [0,1]. $ Prove that $$(x^2+y^2+z^2)(x+y+z-2)\leq 3xyz.$$ p/6625732182

triangle112 wrote: If a b c positive reals smaller than 1, prove: $a+b+c+2abc>ab+bc+ca+2\sqrt{abc}$ It is very nice inequality. After using $c+ab\ge 2\sqrt{abc}$, we need to prove $a+b+c+2abc>ab+bc+ca+(c+ab)\ \ \iff \ \ (1-c)(a+b-2ab)>0$ which is true because $a+b\ge 2\sqrt{ab}>2ab$