Rank–trace inequality: $c\,\mathrm{tr} P - \tfrac{c^2}{4} r + 2c\,\mathrm{tr} Q

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Rank–trace inequality: $c\,\mathrm{tr} P - \tfrac{c^2}{4} r + 2c\,\mathrm{tr} Q - c^2 b \le \|P+Q\|_F^2$ · Prove2Me