Do \(\left\{{}\begin{matrix}a\ge0\\b\ge1\\a+b+c=5\end{matrix}\right.\) \(\Rightarrow c\le4\)

\(\Rightarrow2\le c\le4\Rightarrow\left(c-2\right)\left(c-4\right)\le0\Rightarrow c^2\le6c-8\)

\(0\le a\le1< 6\Rightarrow a\left(a-6\right)\le0\Rightarrow a^2\le6a\)

\(1\le b\le2< 5\Rightarrow\left(b-1\right)\left(b-5\right)\le0\Rightarrow b^2\le6b-5\)

Cộng vế:

\(a^2+b^2+c^2\le6\left(a+b+c\right)-13=17\)

\(A_{max}=17\) khi \(\left(a;b;c\right)=\left(0;1;4\right)\)

\(Q=\dfrac{2-\dfrac{c}{a}-\dfrac{2b}{a}+\left(\dfrac{b}{a}\right)\left(\dfrac{c}{a}\right)}{1-\dfrac{b}{a}+\dfrac{c}{a}}=\dfrac{2-mn+2\left(m+n\right)-mn\left(m+n\right)}{1+m+n+mn}\)

\(Q=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{\left(m+1\right)\left(n+1\right)}\ge\dfrac{\left[8-\left(m+n\right)^2\right]\left(m+n+1\right)}{\left(m+n+2\right)^2}\)

Đặt \(m+n=t\Rightarrow0\le t\le2\)

\(Q\ge\dfrac{\left(8-t^2\right)\left(t+1\right)}{\left(t+2\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{\left(2-t\right)\left(4t^2+15t+10\right)}{4\left(t+2\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu "=" xảy ra khi \(t=2\) hay \(m=n=1\)

Ta có

(a+1)+(b+10)+(c+2014)+(d+2017)\(\le\) 4(d+2017) ( phần này tự lập luận nhé, cũng dễ mà)

=> (a+b+c+d)+(1+10+2014+2017)\(\le\) 4(d+2017)

=> 4042+4042\(\le\) 4(d+2017)

=>8084\(\le\) 4(d+2017)

=> \(2021\le d+2017\)

=> \(4\le d\)

Vậy GTNN của d là 4

k cho mình nhé bạn bạn k mình 1 k mình k bạn 3 k nhé

\(\Leftrightarrow ab^2+bc^2+ca^2\ge a^2b+b^2c+c^2a\)

\(\Leftrightarrow\left(c^2b-abc-b^2c+ab^2\right)+\left(ca^2+abc-ac^2-a^2b\right)\ge0\)

\(\Leftrightarrow b\left(c^2-ac-bc+ab\right)-a\left(c^2-ac-bc+ab\right)\ge0\)

\(\Leftrightarrow\left(b-a\right)\left(c^2-ac-bc+ab\right)\ge0\)

\(\Leftrightarrow\left(b-a\right)\left(c-b\right)\left(c-a\right)\ge0\) (luôn đúng do \(c\ge b\ge a>0\))

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