?????

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(a+b+c+1\right)^2=\left(a.1+\frac{1}{\sqrt{2}}.\sqrt{2}\left(b+c\right)+\frac{1}{\sqrt{2}}.\sqrt{2}\right)^2\)\(\le\left(a^2+1\right)\text{[}3+2\left(b+c\right)^2\text{]}\)

Khi đó cần CM BĐT : \(\frac{5}{16}\text{[}3+2\left(b+c\right)^2\text{]}\le\left(b^2+1\right)\left(c^2+1\right)\)

Hay: \(16b^2c^2+6\left(b^2+c^2\right)+1\ge20ab\)

BĐT trên đúng theo BĐT AM-GM: \(16b^2c^2+1\ge8bc,6\left(b^2+c^2\right)\ge12bc\)

Dấu '=' xảy ra khi và chỉ khi a=b=c=1/2

TA CÓ: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Do đó: \(\frac{a^3+b^3+c^3}{4abc}=\frac{3}{4}+\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{4abc}\)

\(=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(\ge\frac{3}{4}+\frac{1}{4}.\frac{9}{ab+bc+ca}\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{3}{4}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{9}{4}=\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{1}{30}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{9\left(a^2+b^2+c^2\right)}{4\left(ab+bc+ca\right)}-\frac{131\left(a^2+b^2+c^2\right)}{60\left(ab+bc+ca\right)}-\frac{3}{2}\)

\(=\frac{-22}{15}+\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}+\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\)

\(\ge\frac{-22}{15}+2\sqrt{\left[\frac{ab+bc+ca}{15\left(a^2+b^2+c^2\right)}\right]\left[\frac{a^2+b^2+c^2}{15\left(ab+bc+ca\right)}\right]}=\frac{-22}{15}+\frac{2}{15}=\frac{-4}{3}\)

Dấu '=' xảy ra <=> a=b=c

Vậy GTNN của P là -4/3 khi a=b=c

bạch cầu  giúp tránh nhiễm trùng 

hồng cầu giúp lưu thông máu 

tiểu cầu giúp máu đông lại

x( x² - y ) - x²( x - y ) + 1817

= x³ - xy - x³ + x²y + 1817

= x²y - xy + 1817

Thế x = -1 ; y = 100 ta được :

(-1)².100 - (-1).100 + 1817

= 100 + 100 + 1817

= 2017

HELP ME PLS :<