ủng hộ mk nha mọi người

mình đag gấp nhờ mọi người giải giúp

"Chấm" nhẹ hóng cao nhân ạ :)

P/s: mong các bác giải theo cách lớp 8 ạ :) Tặng 5SP / 1 câu nhé ;)

Câu 3: Tham khảo đây nhá: Câu hỏi của Trương Thanh Nhân, t làm r,giờ lười đánh lại.

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

<=>\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)

<=>\(ab+bc+ca=0\)

<=>\(\frac{ab+bc+ca}{abc}=0\)

<=> \(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)

<=>\(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

<=>\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c}^3\)

<=>\(\frac{1}{a^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{b^3}=\frac{-1}{c}^3\)

<=>\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Ta có: \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=\frac{3abc}{abc}=3\)

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

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

vay la song cau nhe

a^2+c^2=bc+c^2=c(b+c)

b^2+a^2=b^2+bc=b(b+c)

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