Typical Inequalities for real numbers

Question

If a,b,c and d are real numbers, then

(a) $\bigg(\frac{5a}{12}+\frac{b}{3}+\frac{c}{6}+\frac{d}{12}\bigg)^2 \leq \frac{5a^2}{12}+\frac{b^2}{3}+\frac{c^2}{6}+\frac{d^2}{12}$

(b) $a+b+c=2$ with $0<a,b,c<1$ implies that $\frac{a}{1-a}.\frac{b}{1-b}.\frac{c}{1-c} \geq 8$

Please give me some hint to prove these inequalities. Thanks in advance.

Answer 1

$(2)$ We can write $$\displaystyle a= \frac{(a+b-c)+(a-b+c)}{2}\geq \sqrt{(a+b-c)\cdot (a-b+c)}$$

Similarly $$\displaystyle b = \frac{(b+c-a)+(b-c+a)}{2}\geq \sqrt{(b+c-a)\cdot (b-c+a)}$$

Similarly $$\displaystyle c = \frac{(c+a-b)+(c-a+b)}{2}\geq \sqrt{(c+a-b)\cdot (c-a+b)}$$

So we get $$\displaystyle abc \geq (b+c-a)\cdot (c+a-b)\cdot (a+b-c) = (2-2a)(2-2b)(2-2c)$$

above we have used $$a+b+c = 2$$

so we get $$\displaystyle \frac{abc}{(1-a)(1-b)(1-c)}\geq 8$$

Answer 2

For (a), note that $5/12+1/3+1/6+1/12=1$, so the result follows from the convexity of the function $f(x)=x^2$.

Answer 3

(a). Direct consequence of Cauchy-Schwarz inequality.

$$\bigg(\frac{5a}{12}+\frac{b}{3}+\frac{c}{6}+\frac{d}{12}\bigg)^2 \leq \left(\frac{5a^2}{12}+\frac{b^2}{3}+\frac{c^2}{6}+\frac{d^2}{12}\right)\left(\frac{5}{12}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}\right)$$

Answer 4

for(b):

let $ x=\dfrac{a}{1-a},y=\dfrac{b}{1-b},z=\dfrac{c}{1-c} \implies a=\dfrac{x}{1+x}, b=\dfrac{y}{1+y}, c=\dfrac{z}{1+z} \\ a+b+c=2 \implies xyz=2+x+y+z \ge 2+3\sqrt[3]{xyz}\\ \sqrt[3]{xyz}=u \implies u^3\ge 2+3u \implies u \ge 2 \implies xyz \ge 8$