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Lời giải tại link sau:

https://hoc24.vn/cau-hoi/cho-abc-la-cac-so-duongcmr-dfrac1a2bcdfrac1b2acdfrac1c2abledfracabc2abc.193908584039

\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

Xét \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\Leftrightarrow ab+bc+ac=abc\)

\(\Rightarrow VT=\dfrac{a^3}{a^2+ab+bc+ac}+\dfrac{b^3}{b^2+ab+bc+ac}+\dfrac{c^3}{c^2+ab+bc+ac}\)

\(\Leftrightarrow VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(b+a\right)\left(b+c\right)}+\dfrac{c^3}{\left(c+b\right)\left(c+a\right)}\)

Áp dụng bđt Cauchy ta có :

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge3\sqrt[3]{\dfrac{a^3}{64}}=\dfrac{3a}{4}\)

Thiết lập tương tự và thu lại ta có :

\(VT+\dfrac{a+b+c}{2}\ge\dfrac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(a+b+c\right)-\dfrac{1}{2}\left(a+b+c\right)=\dfrac{a+b+c}{4}\left(đpcm\right)\)

Dấu '' = '' xảy ra khi \(a=b=c=3\)

a/ \(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ac+a^2}\)

\(=\dfrac{a^4}{a^3+a^2b+ab^2}+\dfrac{b^4}{b^3+b^2c+bc^2}+\dfrac{c^4}{c^3+ac^2+ca^2}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a\left(a^2+ab+b^2\right)+b\left(b^2+bc+c^2\right)+c\left(c^2+ca+a^2\right)}\)

\(=\dfrac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}=\dfrac{a^2+b^2+c^2}{a+b+c}\)

b/ \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}=\dfrac{a^4}{abc}+\dfrac{b^4}{abc}+\dfrac{c^4}{abc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3abc}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}\)

\(\ge\dfrac{3\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)\left(a+b+c\right)}=\dfrac{3\left(a^2+b^2+c^2\right)^2}{a+b+c}\)

Ta có

\(\sum\dfrac{a}{a+\sqrt{2019a+bc}}=\sum\dfrac{a}{a+\sqrt{a^2+a\left(b+c\right)+bc}}\)

Áp dụng AM - GM : \(b+c\ge2\sqrt{bc}\)

\(\Rightarrow\sum\dfrac{a}{a+\sqrt{a^2+a\left(b+c\right)+bc}}\le\dfrac{a}{a+\sqrt{a^2+2a\sqrt{bc}+bc}}\)

\(=\sum\dfrac{a}{a+\sqrt{\left(a+\sqrt{bc}\right)^2}}=\sum\dfrac{a}{a+a+\sqrt{bc}}\)

Tự làm tiếp

Lời giải:

Áp dụng BĐT Schur bậc 3:

\(abc\geq (a+b-c)(b+c-a)(c+a-b)\)

\(\Leftrightarrow abc\geq (1-2c)(1-2a)(1-2b)\)

\(\Leftrightarrow 9abc\geq 4(ab+bc+ac)-1\) (thay \(a+b+c=1\) )

\(\Rightarrow abc\geq \frac{4}{9}(ab+bc+ac)-\frac{1}{9}\)

\(\Rightarrow ab+bc+ac-abc\leq \frac{5}{9}(ab+bc+ac)+\frac{1}{9}\)

Theo hệ quả quen thuộc của BĐT AM-GM

\(ab+bc+ac\leq \frac{(a+b+c)^2}{3}=\frac{1}{3}\Rightarrow \frac{5}{9}(ab+bc+ac)+\frac{1}{9}\leq \frac{8}{27}\)

\(\Rightarrow ab+bc+ac-abc\leq \frac{8}{27}\) (đpcm)

Dấu bằng xảy ra khi $3a=3b=3c=1$

À rồi, vừa mới nghĩ ra cách khác:

Ta có:

\(ab+bc+ac-abc=b(a+c)+ac-ac(1-a-c)\)

\(=b(a+c)+ac(a+c)=(a+c)(b+ac)=(a+c)(1-a-c+ac)=(a+c)(1-a)(1-c)\)

Áp dụng BĐT AM-GM:
\((a+c)(1-a)(1-c)\leq \left(\frac{a+c+1-a+1-c}{3}\right)^3=\frac{8}{27}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$

Áp dụng bất đẳng thức AM-GM ta có :

\(ab\le\frac{\left(a+b\right)^2}{4}\)=> \(\frac{ab}{a+b}\le\frac{a+b}{4}\)

Tương tự : \(\frac{bc}{b+c}\le\frac{b+c}{4}\); \(\frac{ca}{c+a}\le\frac{c+a}{4}\)

Cộng vế với vế các bđt trên ta có đpcm

Đẳng thức xảy ra <=> a=b=c

Có     ab\le\dfrac{\left(a+b\right)^2}{4}\Rightarrow\dfrac{ab}{a+b}\le\dfrac{a+b}{4}ab≤4(a+b)2​⇒a+bab​≤4a+b​.

Lời giải:

Áp dụng BĐT AM-GM:

\(\text{VT}\leq \frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

\(\leq \frac{\frac{a+b}{2}+\frac{b+c}{2}+\frac{a+c}{2}}{2abc}=\frac{a+b+c}{2abc}=\text{VP}\)

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$